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MACHINE DESIGN 



BY 



ALBERT W. SMITH 
• i 

Director of Sibley College, Cornell University 



GUIDO H. MARX 

Professor of Machine Design 
Leland Stanford Junior University 



FOURTH EDITION, REVISED AND ENLARGED 



NEW YORK 
JOHN WILEY & SONS, Inc. 
London: CHAPMAN & HALL, Limited 
i9 J 5 






Copyright, 190S, 1908, 1909, 191S, 

BY 

ALBERT W. SMITH and GUI DO H. MARX 



1$ 



j/w/t. 



THE SCIENTIFIC PRESS 

ROBERT DRUMMOND AND COMPANY 

BROOKLYN. N. Y. 



g)C!,A410984 



PREFACE TO FOURTH EDITION. 



The call for a new edition of this book has given opportunity 
for thorough revision of the text and the inclusion of results of 
recent investigations of machine elements. The task, by cordial 
mutual consent, was undertaken and has been executed solely 
by the junior author, upon whom the entire responsibility for the 
book now rests. 

The original plan of emphasizing fundamental principles 
and methods of reasoning is retained and there has been no 
effort to make the volume cyclopaedic in scope. 

Engineering literature has been freely consulted and the in- 
debtedness to other writers is acknowledged in the text, as in the 
earlier editions. Similarly, references are made to more exhaust- 
ive treatments of many topics than are possible in this volume. 

Thanks are extended to Assistant Professor L. E. Cutter, 
Mr. B. M. Green, graduate assistant in machine design, and Mr. 
D. J. Conant, all of Stanford University, for the preparation of 
many of the drawings for the new illustrations. 

iii 



PREFACE TO THE SECOND EDITION. 



One can never become a machine designer by studying books. 
Much help may come from books, but the true designer must 
have judgment, ripened by experience, in constructing and 
operating machines. One may know the laws that govern the 
development, transmission and application of energy; may have 
knowledge of constructive materials; may know how to obtain 
results by mathematical processes, and yet be unable to design 
a good machine. There is also needed a knowledge of many 
things connected with manufacture, transportation, erection and 
operation. With this knowledge it is possible to take results 
of computation and accept, reject and modify until a machine 
is produced that will do the required work satisfactorily. 

Professor John E. Sweet once said, "It is comparatively easy 
to design a good new machine, but it is very hard to design a 
machine that will be good when it is old." A machine must 
not only do its work at first, but must continue to do it with a 
minimum of repairs as long as the work needs to be done. The 
designer must be able to foresee the results of machine operation; 
he must have imagination. This is an inborn power, but it may 
be developed by use and by engineering experience. 

But there is a certain part of the designer's mental equipment 
that may be furnished in the class-room, or by books. This is 
the excuse for the following pages. Machine design cannot be 
treated exhaustively. There are too many kinds of machines 
for this and their differences are too great. In this book an 



vi PREFACE. 

effort is made simply to give principles that underlie all machine 
design and to suggest methods of reasoning which may be helpful 
in the designing of any machine. A knowledge of the usual 
university course in pure and applied mathematics is pre- 
supposed. 



CONTENTS. 



CHAPTER I. 

PAGE 

Preliminary i 

CHAPTER II. 
Motion in Mechanisms x 5 

CHAPTER III. 
Parallel or Straight-line Motions 41 

CHAPTER IV. 
Cams 5I 

CHAPTER V. 
Energy in Machines 62 

CHAPTER VI. 
Proportions of Machine Parts as Dictated by Stress 81 

CHAPTER VII. 
Riveted Joints 105 

CHAPTER VIII. 
Bolts and Screws 139 

CHAPTER IX. 

Means for Preventing Relative Rotation 168 

vii 



viii CONTENTS. 

CHAPTER X. 

PAGE 

Sliding Surfaces • • 185 

CHAPTER XL 
Axles, Shafts, and Spindles 194 

CHAPTER XII. 
Journals, Bearings, and Lubrication 210 

CHAPTER XIII, 
Roller- and Ball-bearings 261 

CHAPTER XIV. 
Couplings and Clutches 274 

CHAPTER XV. 
Belts, Ropes, Brakes, and Chains - • 286 

CHAPTER XVI. 
Fly-wheels and Pulleys • • 328 

CHAPTER XVII. 
v Toothed Wheels or Gears 347 

CHAPTER XVIII. 
Springs • • . . 429 

CHAPTER XIX. 
Machine Supports 436 

CHAPTER XX. 
Machine Frames 441 



Appendix 477 

Index 4 8 5 



INTRODUCTION. 



In general there are five considerations of prime importance 
in designing machines: I. Adaptation, II. Strength and Stiff- 
ness, III. Economy, IV. Appearance, V. Safety. 

I. This requires all complexity to be reduced to its lowest 
terms in order that the machine shall accomplish the desired 
result in the most direct way possible, and with greatest convenience 
to the operator. 

II. This requires the machine parts subjected to the action of 
forces to sustain these forces, not only without rupture, but also 
without such yielding as would interfere with the accurate action 
of the machine. In many cases the forces to be resisted may 
be calculated, and the laws of mechanics and the known qualities 
of constructive materials become factors in determining propor- 
tions. In other cases the force, by the use of a "breaking-piece," 
may be limited to a maximum value, which therefore dictates 
the design. But in many other cases the forces acting are neces- 
sarily unknown; and appeal must be made to the precedent of 
successful practice, or to the judgment of some experienced man, 
until one's own judgment becomes trustworthy by experience. 

In proportioning machine parts, the designer must always be 
sure that the stress which is the basis of the calculation or the 
estimate, is the maximum possible stress; otherwise the part 
will be incorrectly proportioned. For instance, if the arms of a 
pulley were to be designed solely on the assumption that they 



x INTRODUCTION. 

endure only the transverse stress due to the belt tension, they 
would be found to be absurdly small, because the stresses resulting 
from the shrinkage of the casting in cooling are often far greater 
than those due to the belt pull. 

The design of many machines is a result of what may be called 
"machine evolution." The first machine was built according to 
the best judgment of its designer; but that judgment was fallible, 
and some part ruptured under the stresses sustained; it was re- 
placed by a new part made stronger; it ruptured again, and again 
was enlarged, or perhaps made of some more suitable material; 
it then sustained the applied stresses satisfactorily. Some other 
part yielded too much under stress, although it was entirely safe 
from actual rupture; this part was then stiffened and the process 
continued till the whole machine became properly proportioned 
for the resisting of stress. Many valuable lessons have been learned 
from this process; many excellent machines have resulted from 
it. There are, however, two objections to it : it is slow and very 
expensive, and if any part had originally an excess of material, 
it is not changed; only the parts that yield are perfected. 

Modern analytical methods are rigntly displacing it in all 
progressive establishments. 

III. The attainment of economy does not necessarily mean the 
saving of metal or labor, although it may mean that. To illustrate : 
Suppose that it is required to design an engine -lathe for the 
market. The competition is sharp; the profits are small. How 
shall the designer change the design of the lathes on the market 
to increase profits? (a) He may, if possible, reduce the weight 
of metal used, maintaining strength and stiffness by better dis- 
tribution. But this must not increase labor in the foundry or 
machine-shop, nor reduce weight which prevents undue vibrations. 
(b) He may design special tools to reduce labor without reduction 
of the standard of workmanship. The interest on the first cost 
of these special tools, however, must not exceed the possible gain 



INTRODUCTION. xi 

from increased profits. (c) He may make the lathe more con- 
venient for the workmen. True economy permits some increase 
in cost to gain this end. It is not meant that elaborate and 
expensive devices are to be used, such as often come from men 
of more inventiveness than judgment; but that if the parts can 
be rearranged, or in any way changed, so that the lathes-man 
shall select this lathe to use because it is handier when other 
lathes are available, then economy has been served, even though 
the cost has been somewhat increased, because the favorable 
opinion of intelligent workmen means increased sales. 

In (a) economy is served by a reduction of metal; in (b) by a 
reduction of labor; in (c) it may be served by an increase of both 
labor and material. 

The addition of material largely in excess of that necessary 
for strength and rigidity, to reduce vibrations, may also be in the 
interest of economy, because it may increase the durability of the 
machine and its foundation, or may reduce the expense incident 
upon repairs and delays, thereby bettering the reputation of the 
machine and increasing sales. 

Suppose, to illustrate further, that a machine part is to be 
designed, and either of two forms, A or B, will serve equally well. 
The part is to be of cast iron. The pattern for A will cost twice 
as much as for B. In the foundry and machine-shop, however, 
A can be produced a very little cheaper than B- Clearly then if 
but one machine is to be built, B should be decided on ; whereas, 
if the machine is to be manufactured in large numbers, A is 
preferable. Expense for patterns is a first cost. Expense for 
work in the foundry and machine-shop is repeated with each 
machine. 

Economy of operation also needs attention. This depends 
upon the efficiency of the machine ; i.e., upon the proportion of the 
energy supplied to the machine which really does useful work. 
This efficiency is increased by the reduction of useless frictional 



xii INTRODUCTION. 

resistances, by careful attention to the design and means of lubri- 
cation of rubbing surfaces. 

In order that economy may be best attained, the machine 
designer needs to be familiar with all the processes used in the 
construction of machines — pattern -making, foundry work, forging, 
and the processes of the machine-shop — and must have them con- 
stantly in mind, so that while each part designed is made strong 
enough and stiff enough, and properly and conveniently arranged, 
and of such form as to be satisfactory in appearance, it also is 
so designed that the cost of construction is a minimum. 

IV. The fourth important consideration is Appearance. 
There is a beauty possible of attainment in the design of machines 
which is always the outgrowth of a purpose. Otherwise expressed, 
a machine to be beautiful must be purposeful. Ornament for 
ornament's sake is seldom admissible in machine design. And 
yet the striving for a pleasing effect is as much a part of the duty 
of a machine designer as it is a part of the duty of an architect. 

As a guiding principle, the general rule may be laid down 
that simplicity and directness are always best. Each member 
should be studied with strict reference to the function which it 
is to perform and the stresses to which it is subjected and then 
given the form and size best suited to meet the conditions with 
the greatest economy of material and workmanship. When 
combined, the parts must be modified in such manner as may be 
found necessary to the harmonious effect of the whole. 

V. Safety of the operator and others who come into the 
vicinity of the machine is the fifth important point in design. 
It is really a sub-division of Adaptation but, for emphasis, may 
be given the prominence of a separate head. Beyond the pro- 
visions for Strength and Stiffness, it requires that all moving 
parts shall be so formed and guarded as to eliminate, so far as 
may be foreseen, all danger of bodily accident. 



MACHINE DESIGN 



CHAPTER I. 

PRELIMINARY. 

i. Definitions. — The study of machine design is based upon 
the science of mechanics, which treats questions involving the 
consideration of motion, force, work, and energy. Since it will 
be necessary to use these terms almost continually, it is well to 
make an exact statement of what is to be understood by them. 

Motion may be defined as change of position in space. 

A Force is one of a pair of equal, opposite, and simultaneous 
actions between two bodies by which the state of their motion 
is altered, or a change in the form or condition of the bodies them- 
selves is effected. 

Work is the name given to the result of a force in motion. 

Energy is the capacity possessed by matter to do work. 

A Machine is a combination of resistant bodies whose relative 
motions are completely constrained, and whose function it is to 
transform available energy into useful work. 

The law of Conservation of Energy underlies every machine 
problem. This law may be expressed as follows: The sum of 
energy in the universe is constant. Energy may be transferred 
in space; it may be stored for varying lengths of time; it may 
be changed from one of its several forms to another; but it can- 
not be created or destroyed. 



2 MACHINE DESIGN. 

The application of this law to machines is as follows: A 
machine receives energy from a source, and uses it to do useful 
and useless work. 

A single cycle of action of a machine is that sequence of opera- 
tions during which each member of the machine has gone once 
through all the relative motions possible to it. A complete cycle 
of action is such a period that all conditions (velocities, etc., as 
well as relative positions) in the machine are the same at its 
beginning and end. 

During a single cycle of action of the machine, the energy 
received equals the total work done. The work done may appear 
as (a) useful work delivered by the machine, or as (b) heat due 
to energy transformed through frictional- resistance, or as (c) 
stored mechanical energy in some moving part of the machine 
whose velocity is increased. The sign of the stored energy may 
be plus or minus, so that energy received in one cycle may be 
delivered during another cycle; but for any considerable time 
interval of machine action the algebraic sum of the stored energy 
must equal zero. 

For a single cycle: 
Energy received = useful work + useless work ± stored energy. 

For continuous action: 

Energy received = useful work + useless work. 
In operation a machine generally acts by a continuous repetition 
of its cycle. 

2. Efficiency of Machines. — In general, efficiency may be 
defined as the ratio of a result to the effort made to produce that 
result. In a machine the result corresponds to the useful work, 
while the effort corresponds to the energy received. Hence the 
efficiency of a machine = useful work -^ energy received.* The 
designer must strive for high efficiency, i.e., for the greatest 
possible result for a given effort. 

* The work and energy must, of course, be expressed in the same units. 



PRELIMINARY. 3 

3. Function of Machines. — Nature furnishes sources of 
energy, and the supplying of human needs requires work to be 
done. The function of machines is to cause matter possessing 
energy to do useful work. 

The chief sources of energy in nature available for machine 
purposes are: 

1st. The energy of air in motion {i.e., wind) due to its mass 
and velocity. 

2d. The energy of water due to its mass and motion or posi- 
tion. 

3d. The energy dormant in fuels which manifests itself as 
heat upon combustion. 

The general method by which the machine function is exer- 
cised may be shown by the following illustration: 

Illustration. — The water in a mill-pond possesses energy 
(potential) by virtue of its position. The earth exerts an attrac- 
tive force upon it. If there is no outlet, the earth's attractive 
force cannot cause motion; and hence, since motion is a neces- 
sary factor of work, no work is done. 

If the water overflows the dam, the earth's attraction causes 
that part of it which overflows to move to a lower level, and before 
it can be brought to rest again it does work against the force 
which brings it to rest. If this water simply falls upon rocks, 
its energy is transformed into heat, with no useful result. 

But if the water is led from the pond to a lower level, in a 
closed pipe which connects with a water-wheel, it will act upon 
the vanes of the wheel (because of the earth's attraction), and 
will cause the wheel and its shaft to rotate against resistance, 
whereby it may do useful work. The water-wheel is a machine 
and is called a Prime Mover, because it is the first link in the 
machine-chain between natural energy and useful work, and it 
is the function of prime movers to change some form of natural 
energy into a form applicable to the performance of useful work. 



4 MACHINE DESIGN. 

Since it is usually necessary to do the required work at some 
distance from the necessary location of the water-wheel, Machinery 
of Transmission is used (shafts, pulleys, belts, cables, etc.), and 
the rotative energy is rendered available at the required place.* 

But this rotative energy may not be suitable to do the re- 
quired work; the rotation may be too slow or too fast; a resist- 
ance may need to be overcome in straight, parallel lines, or at 
periodical intervals. Hence Machinery of Application is intro- 
duced to transform the energy to meet the requirements of the 
work to be done. Thus the chain is complete, and the potential 
energy of the water does the required useful work. 

The chain of machines which has the steam-boiler and engine 

for its prime mover transforms the potential heat energy of 

fuel into useful work. This might be analyzed in a similar way. 

4. Free Motion. — The general science of mechanics treats of 

the action of forces upon "free bodies." 

In the case of a lt free body " acted on by a system of forces 
not in equilibrium, motion results in the direction of the resultant 
of the system. If another force is introduced whose line of 
action does not coincide with that of the resultant, the line of 
action of the resultant is changed, and the body moves in a new 
direction. The character of the motion, therefore, is dependent 
upon the forces which produce the motion. 
This is called free motion. 

Example. — In Fig. 1, suppose the free 
body M to be acted on by the concurrent 
forces 1, 2, and 3 whose lines of action 
pass through the center of gravity of M. 
The line of action of the resultant of these 
FlG j forces is AB, and the body's center of 

gravity would move along this line. 

* Electric, hydraulic, and pneumatic transmission systems are also em- 
ployed. 




PRELIMINARY. 5 

If another force, 4, is introduced, CD becomes the line of 
action of the resultant, and the motion of the body is along 
the line CD. 

5. Constrained Motion. — In a machine certain definite 
motions occur; any departure from these motions, or the pro- 
duction of any other motions, would result in derangement 
of the action of the machine. Thus, the spindle of an engine- 
lathe turns accurately about its axis; the cutting-tool moves 
parallel to the spindle's axis; and an accurate cylindrical surface 
is thereby produced. If there were any departure from these 
motions, the lathe would fail to do its required work. In all 
machines certain definite motions must be produced, and all 
other motions must be prevented; or, in other words, motion 
in machines must be constrained. 

Constrained motion differs from free motion in being inde- 
pendent of the forces which produce it. If any force, not suffi- 
ciently great to produce deformation, be applied to a body whose 
motion is constrained, the result is either a certain predeter- 
mined motion, or no motion at all. 

6. Force Opposed by Passive Resistance. — A force may act 
without being able to produce motion (and hence without being 
able to do work), as in the case of the water in a mill-pond without 
overflow or outlet. This may be further illustrated: Suppose a 
force, say hand pressure, to be applied vertically to the top of a 
table. The material of the table offers a passive resistance, and 
the force is unable to produce motion, or to do work. 

It is therefore possible to offer passive resistance to such 
forces as may be required not to produce motion, thereby render- 
ing them incapable of doing work. Whenever a body opposes 
a passive resistance to the action of a force a change in its condi- 
tion is effected: the force sets up an equivalent stress in the 
material of the body. Thus, when the table offers a passive 
resistance to the hand -pressure, compressive stress is induced 



MACHINE DESIGN. 



in the legs. In every case the material of the body must be of such 
shape and strength as to resist successfully the induced stress. 

In a machine there must be provision for resisting every pos- 
sible force which tends to produce any but the required motion. 
This provision is usually made by means of the passive resistance 
of properly formed and sufficiently resistant metallic surfaces. 

Illustration I. — Fig. 2 represents a section and end view of 
a wood-lathe headstock. It is required that the spindle, S, and 
the attached cone pulley, C, shall have no other motion than 




rotation about the axis of the spindle. If any other motion is 
possible, this machine part cannot be used for the required pur- 
pose. At A and B the cylindrical surfaces of the spindle are 
enclosed by accurately fitted bearings or internal cylindrical sur- 
faces. Suppose any force, P, whose line of action lies in the 
plane of the paper, to be applied to the cone pulley. It may be 
resolved into a radial component, R, and a tangential component, 
T. The passive resistance of the cylindrical surfaces of the 
journal and its bearing, prevents R from producing motion; 
while it offers no resistance, friction being disregarded, to the 
action of T, which is allowed to produce the required motion, 
i.e., rotation about the spindle's axis. If the line of action of P 
pass through the axis, its tangential component becomes zero, 



PRELIMINARY. • 7 

and no motion results. If the line of action of P become tangen- 
tial, its radial component becomes zero, and P is wholly applied 
to produce rotation. If a force Q, whose line of action lies in 
the plane of the paper, be applied to the cone, it may be resolved 
into a radial component, N, and a component, M, parallel to 
the spindle's axis. N is resisted as before by the journal and 
bearing surfaces, and M is resisted by the shoulder surfaces of 
the bearings, which fit against the shoulder surfaces of the spindle 
collars. The force Q can therefore produce no motion at all. 

In general, any force applied to the cone pulley may be 
resolved into a radial, a tangential, and an axial component. 
Of these only the tangential component is able to produce motion; 
and that motion is the motion required. The constrainment is 
therefore complete; i.e., there can be no motion except rotation 
about the spindle's axis. This result is due to the passive resist- 
ance of metallic surfaces. 

Illustration II. — R, Fig. 3, represents, with all details omitted, 
the "ram," or portion of a shaping-machine which carries the 




Fig. 3. 

cutting-tool. It is required to produce plane surfaces, and hence 
the "ram" must have accurate rectilinear motion in the direction 
of HK. Any deviation from such motion would render the 
machine useless. 

Consider Fig. 3, A. Any force which can be applied to the 
ram may be resolved into three components: one vertical, one 
horizontal and parallel to the paper, and one perpendicular to 



8 MACHINE DESIGN. 

the paper. The vertical component, if acting upward, is resisted 
by the plane surfaces in contact at C and D ; if acting downward, 
it is resisted by the plane surfaces in contact at E. Therefore 
no vertical component can produce motion. The horizontal 
component parallel to the paper is resisted by the plane surfaces 
in contact at F or G, according as it acts toward the right or 
left. The component perpendicular to the paper is free to pro- 
duce motion in the direction of its line of action; but this is the 
motion required. 

Any force, therefore, which has a component perpendicular 
to the paper can produce the required motion, but no other 
motion. The constrainment is therefore complete, and the 
result is due to the passive resistance offered by metallic surfaces. 

Complete Constrainment is not always required in machines. 
It is only necessary to prevent such motions as interfere with 
the accomplishment of the desired result. 

The weight of a moving part is sometimes utilized to produce 
constrainment in one direction. Thus in a planer-table, and in 
some lathe -carriages, downward motion and unallowable side 
motion are resisted by metallic surfaces; while upward motion 
is resisted by the weight of the moving part. 

From the foregoing it follows that, as passive resistances 
can be opposed to all forces whose lines of action do not coincide 
with the desired direction of motion of any machine part, it may 
be said that the nature of the motion is independent of the forces 
producing it. 

Since the motions of machine parts are independent of the forces 
producing them, it follows that the relation of such motions may 
be determined without bringing force into the consideration. 

7. Kinds of Motion in Machines. — Motion in machines may 
be very complex, but it is chiefly plane motion. 

When a body moves in such a way that any section of it re- 
mains in the same plane, its motion is called plane motion. All 



PRELIMINARY. Q 

sections parallel to the above section must also remain, each in 
its own plane. If the plane motion is such that all points of the 
moving body remain at a constant distance from some line, AB, 
the motion is called rotation about the axis AB. Example. — 
A line-shaft with attached parts. 

If all points of a body move in straight parallel paths, the 
motion of the body is called rectilinear translation. Examples. — 
Engine cross-head, lathe -carriage, planer-table, shaper-ram. 
Rectilinear translation may be conveniently considered as a 
special case of rotation, in which the axis of rotation is at an 
infinite distance, at right angles to the motion. 

If a body moves parallel to an axis about which it rotates, 
the body is said to have helical or screw motion. Example. — 
A nut turning upon a stationary screw. 

If all points of a body, whose motion is not plane motion, 
move so that their distances from a certain point, O, remain 
constant, the motion is called spheric motion. This is because 
each point moves in the surface of a sphere whose center is O. 
Example. — The arms of a fly-ball steam-engine governor, when 
the vertical position is changing. 

8. Relative Motion. — The motion of any machine part, like 
all known motion, is relative motion. It is studied by reference 
to some other part of the same machine. Some one part of a 
machine is usually (though not necessarily) fixed, i.e., it has no 
motion relative to the earth. This fixed part is called the frame 
of the machine. The motion of a machine part may be referred 
to the frame, or, as is often necessary, to some other part which 
also has motion relative to the frame. 

The kind and amount of relative motion of a machine part 
depend upon the part to which its motion is referred. Since this 
is so, it is always essential, when dealing with the motion of a 
machine part, to specify clearly the standard relative to which, 
for the time being, this motion is being considered. 



IO 



MACHINE DESIGN. 



Illustration.— Fig. 4 shows a press. A is the frame; C is a 
plate which is so constrained that, its llllilll 11111111111 

motion being referred to A, it may 
move vertically, but cannot rotate. 
Motion ol rotation is communicated to 
the screw B. The motion of B re- 
ferred to A is helical motion, i.e., 
combined rotation and translation. 
C, however, shares the translation of 
B, and hence there is left only rotation 
as the relative motion of B and C. Fig. 4. 

The motion of B referred to C is rotation. The motion of C re- 
ferred to B is rotation. The motion of C referred to A is translation. 

In general, if two machine members, M and N, move relative 
to a third member, R, the relative motion of M referred to N 
depends on how much of the motion of N is shared by M . If 
M and N have the same motions relative to R, they have no mo- 
tion relative to each other. 

Conversely, if two bodies have no relative motion, they have 
the same motion relative to a third body. Thus in Fig. 4, if 
the constrainment of C were such that it could share B's rotation, 
as well as its translation, then C would have helical motion rela- 
tive to the frame, and no motion at all relative to B. This is 
assumed to be self-evident. 

A rigid body is one in which the distance between elementary 
portions * is constant. No body is absolutely rigid, but usually 
in machine members the departure from rigidity is so slight that 
it may be neglected. 

Many machine members, as springs, etc., are useful because 
of their lack of rigidity. 



* In this volume this term is used as interchangeable with the term " par- 
ticles " of mechanics. 




PRELIMINARY. II 

Points * in a rigid body can have no relative motion, and 
hence must all have the same motion. 

9. Instantaneous Plane Motion and Instantaneous Centers 
or Centros. — Points of a moving body trace more or less complex 
paths. If a point be considered as 
moving from one position in its path to 
another indefinitely near, its motion is 
called instantaneous motion. The point 
is moving, for the instant, along a 
straight line joining the two indefinitely 
near together positions, and such a 
line is a tangent to the path. In problems which are solved 
by the aid of the conception of instantaneous motion it is only 
necessary to know the direction of motion; hence, for such pur- 
poses, the instantaneous motion of a point is fully defined by a 
tangent to its path at the position occupied by the point at the 
instant. 

Thus in Fig. 5, if a point is moving in the path APB, when 
it occupies the position P the tangent TT represents its instan- 
taneous motion. Any number of curves could be drawn tangent 
to TT at P, and any one of them would be a possible path of 
the point; but whatever path it is following, its instantaneous 
motion is represented by TT. The instantaneous motion of a 
point is therefore independent of the form of its path. Any one 
of the possible paths may be considered as equivalent, for the 
instant, to a circle whose center is anywhere in the normal NN. 

In general, the instantaneous motion of a point, P, is equiva- 
lent to rotation about some point, O, in a line through the point P 
perpendicular to the direction of its instantaneous motion. 

Let the instantaneous motion of a point, A, Fig. 6, in a sec- 
tion of a moving body be given by the line TT. Then the motion 

* Tn this volume this term is used as interchangeable with the term " par- 
ticles " of mechanics. 



12 MACHINE DESIGN. 

is equivalent to rotation about some point on the line i5 as a 

center, but it may be any point, and hence the instantaneous 

motion of the body is not determined. But if the instantaneous 

motion of another point, C, be given by the line T1T1, this motion 

is equivalent to rotation about some point of CD. But the points 

A and C are points in a rigid body, and can have no relative 

motion, and must have the same motion, i.e., rotation about the 

same center. A rotates about some point of AB, and C rotates 

about some point of CD] but they must rotate about the same 

point, and the only point which 

is at the same time in both lines 

is their intersection, O. Hence 

A and C, and all other points 

of the body, rotate, for the instant, 

about an axis of which O is the 

projection; or, in other words, the 

instantaneous motion of the body FlG 6 

is rotation about an axis of which 

O is the projection. This axis is the instantaneous axis of 

the body's motion, and O is the instantaneous center of the 

motion of the section shown in Fig. 6. 

For the sake of brevity an instantaneous center will be called 
a centre 

If TT and T\T\ had been parallel to each other, AB and 
CD would also have been parallel, and would have intersected 
at infinity; in which case the body's instantaneous motion would 
have been rotation about an axis infinitely distant; i.e., it would 
have been translation. 

The motion of the body in Fig. 6 is of course referred to a 
fixed body, which, in this case, may be represented by the paper. 
The instantaneous motion of the body relative to the paper is 
rotation about O. Let M represent the figure, and N the fixed 
body represented by the paper. Suppose the material of M 




PRELIMINARY. 13 

to be extended so as to include O. Then a pin could be put 
through O, materially connecting M and N, without interfering 
with their instantaneous motion Such connection at any other 
point would interfere with the instantaneous motion. 

The centro of the relative motion of two bodies is a point, and 
the only one, at which they have no relative motion; it is a point, 
and the only one, that is common to the two bodies for the instant, 
and which may be considered as being a point of either; it is a 
point, and the only one, about which as center either body may 
be considered as rotating, for the instant, relative to the other. 

It will be seen that the points of the figure in Fig. 6 might 
be moving in any paths, so long as those paths are tangent at 
the points to the lines representing the instantaneous motion. 

In general, centros of the relative motion of two bodies are 
continually changing their position. They may, however, remain 
stationary; i.e., they may become fixed centers of rotation. 

10. Loci of Centros, or Centrodes.* — As centros change posi- 
tion they describe curves of some kind, and these loci of centros 
may be called centrodes. 

Suppose a section of any body, M, to have motion relatively 
to a section of another body, N (fixed), in the same or a parallel 
plane. Centros may be found for a series of positions, and a 
curve drawn through them on the plane of N would be the 
centrode of the motion of M relatively to N. If, now, M be 
fixed and N moved so that the relative motion is the same as 
before, the centrode of the motion of N relatively to M may be 
located upon the plane of M. Each centrode being the locus 
of the centros in its plane, the two centrodes would always have 
one point in common, that point being the centro of the relative 
motion of the two bodies at the particular instant. (Otherwise 

* Centrode is here used in preference to " centroid," proposed by Professor 
Kennedy, because the latter term has grown to be generally accepted in mathe- 
matics as synonymous with " center of mass." 



14 MACHINE DESIGN. 

expressed, the corresponding centrodes are two curves simul- 
taneously generated in the two planes by a common tracing 
point.) Now, since the centro of the relative motion of two 
bodies is a point at which they have no relative motion, and 
since the points of the centrodes become successively the cen- 
tros of the relative motion, it follows that as the motion goes 
on, the centrodes would roll upon each other without slipping. 
Therefore, if the centrodes are drawn, and rolled upon each 
other without slipping, the bodies M and N will have the same 
relative motion as before. From this it follows that the rel- 
ative plane motion of two bodies may be reproduced by rolling 
together, without slipping, the centrodes of that motion. 

ii. Pairs of Motion Elements. — The external and internal 
surfaces by which motion is constrained, as in Figs. 2 and 3, may 
be called pairs of motion elements. The pair in Fig. 2 is called 
a turning pair, and the pair in Fig. 3 is called a sliding pair. 

The helical surfaces by which a nut and screw engage with 
each other are called a twisting pair. These three pairs of 
motion elements have their surfaces m contact throughout. They 
are called lower pairs. Another class, called higher pairs, have 
contact only along elements of their surfaces. Examples. — Cams 
and toothed wheels. 



CHAPTER II. 

MOTION IN MECHANISMS. 

12. Linkages or Motion Chains; Mechanisms. 

In Fig. 7, b is joined to c by a turning pair; 
c " d " sliding " 

d " a " turning " 

a " b " 




Fig. 7. 

Evidently there is complete constrainment of the relative 
motion of a, b, c, and d. For, d being fixed, if any motion occurs 
in either a, b, or c, the other two must have a predetermined 
corresponding motion. 

c may represent the cross-head, b the connecting-rod, and a 
the crank of a steam-engine of the ordinary type. If c were 
rigidly attached to a piston upon which the expansive force of 
steam acts toward the right, a must rotate about ad. This 
r presents a machine. The members a, b, c, and d may be 
represented for the study of relative motions by the diagram, 
Fig. 8. 

This assemblage of bodies, connected so that there is complete 
constrainment of motion, may be called a motion chain or linkage, 

is 



i6 



MACHINE DESIGN. 



and the connected bodies may be called links. The chain 
shown is a simple chain, because no link is joined to more 
than two others. If any links of a chain are joined to more 
than two others, the chain is a compound chain. Examples will 
be given later. 

When one link of a chain is fixed, i.e., when it becomes the 
standard to which the motion of the others is referred, the chain 
is called a mechanism. Fixing different links of a chain gives 
different mechanisms. Thus in Fig. 8, if d is fixed, the mechanism 
is that which is used in the usual type of steam-engine, as in 
Fig. 7. It is called the slider-crank mechanism. 

But if a is fixed, the result is an entirely different mechanism ; 
for b would then rotate about the permanent center ab, d would 
rotate about the permanent center ad, while c would have a more 
complex motion, rotating about a constantly changing centro, 
whose path may be found. 




Fig. 8. 



Fixing b or c would give, in each case, a still different mechan- 



ism. 



13. Location of Centros.— In Fig. 8, d is fixed and it is re- 
quired to find the centers of rotation, either permanent or in- 



MOTION IN MECHANISMS. 1 7 

stantaneous, of the other three links. The motion of a, relative 
to the fixed link d, is rotation about ad, the axis of the turning 
pair joining a and d. When two links are joined by a turning 
pair their centro is always at the axis of the pair. When two 
links are joined by a sliding pair their centro lies at infinity in 
a direction normal to their relative motion of translation. The 
motion of c relative to d is translation, or rotation about a centro 
cd, at infinity vertically. The centro ab lies at the axis of the 
turning pair joining a and b. This point may be considered 
as a point in a or b; in either case it can have but one direction 
of instantaneous motion relative to any one standard. As a 
point in a its motion, relative to d, is rotation about ad. For 
the instant, then, it is moving along a tangent to the circle 
through ab. But, as a point in b, its direction of instantaneous 
motion relative to d must be the same, and hence its motion 
must be rotation about some point in the line ad-ab, extended 
if necessary. Also, b has a point, be, in common with c; and 
by the same reasoning as above be, as a point in b, rotates for 
the instant about some point of the vertical line through be. 
Now ab and be are points of a rigid body, and one rotates for 
the, instant about some point of AB, and the other rotates for 
the instant about some point of CD; hence both ab and be (as 
well as all other points of b) must rotate about the intersection 
of AB and CD. Hence bd is the centro of the motion of b 
relative to d. 

The motion of a may be referred to c (fixed), and ac will be 
found (by reasoning like that applied to b) to lie at the inter- 
section of the lines EF and GH. 

The motion chain in Fig. 8, as before stated, is called the 
slider-crank chain. 

14. Centros of the Relative Motion of Three Bodies are always 
in the Same Straight Line. — In Fig. 8 it will be seen that the 
three centros of any three links lie in the same straight line. 



1 8 MACHINE DESIGN. 

Thus ad, ab, and bd are the centros of the links a, b, and d. This 
is true of any other set of three links. 

Proof. — Let a, b, and c, Fig. 8.4, be any three bodies having 
relative plane motion. Consider a fixed. There will be a 
centra, ab, of the relative motion of b and a; likewise there will 
be a centra, ac, of the relative motion of c and a. Let their 
positions be assumed as represented. There will also be a centra 
be and it must lie either on the line joining ab and ac or off that 
line. Assume it to lie off the line, as shown. 




Fig. 8A. 

By definition it is a point of both b and c, but as a point of 
either it must have the same instantaneous motion relative to 
any other link, such as a. If be lie off the line ab-ac as shown, 
as a point of b relatively to a it has the instantaneous motion 
M, normal to the line joining it to ab. As a point of c relatively 
to a it has the instantaneous motion N, normal to the line joining 
it to ac. But by the definition M and N, being the instantaneous 
motion of the same point relative to a, must coincide. In such 
case the normal lines bc-ab and bc-ac must coincide. No loca- 
tion of be off the line ab-ac can, therefore, fulfill the conditions 
of the definition; they can only be fulfilled by a location on the 
line ab-ac. Hence it may be stated: The three centros of any 
three bodies having relative plane motion must lie on a straight 



MOTION IN MECHANISMS. 19 

line. This important proposition is called Kennedy's Theorem, 
after its discoverer. 

15. Lever-crank Chain. Location of Centros. — Fig. 9 shows 
a chain of four links of unequal length joined to each other by 



X Ld 




: \ 




; \ 




B / \ , 




\ A 




-Jbc \ 




/S^Cx 




c \"*A 


cToT^n 


/ \ / 
fed d I 


13_ - "-» 


1 \ 


ad 1 


\ 


/ 


\ 


/ 




^^ ^ 



Fig. 9. 

turning pairs. The centros ab, . ad, cd, and be may be located at 
once, since they are at the centers of turning pairs which join 
adjacent links to each other. The centros of the relative motion 
of b, c, and d are be, cd, and bd; and these must be in the 
same straight line. Hence bd is in the line B. The centros 
of the relative motion of a, b, and d are ab, bd, and ad; 
and these also must lie in a straight line. Hence bd is in 
the line A. Being at the same time in A and B, it must be 
at their intersection. By employing the same method ac may 
be found. 

16. The Constrainment of Motion in a linkage is inde- 
pendent of the size of the motion elements. As long as the 
cylindrical surfaces of turning pairs have their axes unchanged, 
the surfaces themselves may be of any size whatever, and the 
motion is unchanged. The same is true of sliding and twisting 
pairs. 

In Fig. 10, suppose the turning pair connecting c and d to be 
enlarged so that it includes be. The link c now becomes a 
cylinder, turning in a ring attached to, and forming part of, 



20 



MACHINE DESIGN. 



the link d. be becomes a pin made fast in c and engaging with 
an eye at the end of b. The centres are the same as before 




Fig. io. 



the enlargement of the pair cd, and hence the relative motion 
is the same. 

In Fig. ii the circular portion immediately surrounding cd 
is attached to d. The link c now becomes a ring moving in a 
circular slot. This may be simplified as in Fig. 12, whence c 
becomes a curved block moving in a limited circular slot in d. 
The centres remain as before, the relative motion is the same, 
and the linkage is essentially unchanged. 

If, in the slider-crank mechanism, the turning pair whose 
axis is ah be enlarged till ad is included, as in Fig. 13, the motion 
of the mechanism is unchanged, but the link a is now called 
an eccentric instead of a crank. This mechanism is usually 
used to communicate motion from the main shaft of a steam- 
engine to the valve. It is used because it may be put on the 
main shaft anywhere without interfering with its continuity and 
strength. 

17. Slotted Cross-head. — The mechanism shown in Fig. 14 
is called the " slotted cross-head mechanism" Its centros may 
be found from principles already given. 



MOTION IN MECHANISMS. 



21 



This mechanism is often used as follows: One end of c, as 
E, is attached to a piston working in a cylinder attached to d. 
This piston is caused to reciprocate by the expansive force of 
steam or some other fluid. The other end of c is attached to 



Fig 




Fig. 



another piston, which also works in a cylinder attached to d. 
This piston may pump water or compress gas (for example 
small ammonia compressors for refrigerating plants). The 
crank a is attached to a shaft, the projection of whose axis is 
ad. This shaft also carries a fly-wheel which insures approxi- 
mately uniform rotation. 



22 



MACHINE DESIGN. 



18. Location of Centros in a Compound Mechanism. — It is 

required to find the centros of the compound linkage, Fig. 15. 
In any linkage, each link has a centro relatively to every other 




Fig. 13. 

link; hence, if the number of links = », the number of centros = 
n{n — \). But the centro ab is the same as ba; i.e., each centro 



e£ 



b 



ad I 



3 



Fig. 14. 

is double. Hence the number of centros to be located for any 

n(n — 1) 
linkage = . In the linkage Fig. 15, the number of centros 



6X5 



IS 



* The links are a, b, c, d, e, and /. 
The centros: ab be cd de ef 
ac bd ce df 
ad be cj 
ae bf 
a} 



MOTION IN MECHANISMS. 



2 3 



The portion above the link d is a slider- crank chain, and 
the character of its motion is in no way affected by the attachment 
of the part below d. On the other hand, the lower part is a 
lever-crank chain, and the character of its motion is not affected 
by its attachment to the upper part. The chain may therefore 
be treated in two parts, and the centros of each part may be 
located from what has preceded. Each part will have six centros, 
and twelve would thus be located, ad, however, is common to 




ae ^-.j^-— 



Fig. 15. 



the two parts, and hence only eleven are really found. Four 
centros, therefore, remain to be located. They are be, c), bf, and 
ce. To locate be, consider the three links a, b, and e, and it 
follows that be is in the line A passing through ab and ae; con- 
sidering b, d, and e, it follows that be is in the line B through bd 
and de. Hence be is at the intersection of A and B. Similar 
methods locate the other centros. 

In general, for finding the centros of a compound linkage of 



24 MACHINE DESIGN. 

six links, consider the linkage to be made up of two simple chains, 
and find their centros independently of each other. Then take 
the two links whose centro is required, together with one of 
the links carrying three motion elements (as a, Fig. 15). The 
centros of these links locate a straight line, A, which contains 
the required centro. Then take the two links whose centro is 
required, together with the other link which carries three motion 
elements. A straight line, B, is thereby located, which contains 
the required centro, and the latter is therefore at the intersection 
of A and B. 

19. Velocity is the rate of motion, or motion per unit time. 

Linear velocity is linear space moved through in unit time; 
it may be expressed in any units of length and time; as, miles 
per hour, feet per minute or per second, etc. 

Angular velocity is angular space moved through in unit time. 
In machines, angular velocity is usually expressed in revolutions 
per minute or per second. 

The linear space described by a point in a rotating body, or 
its linear velocity, is directly proportional to its radius, or its 
distance from the axis of rotation. This is true because arcs 
are proportional to radii. 

If A and B are two points in a rotating body, and if Y\ and r 2 
are their radii, then the ratio of linear velocities 

linear veloc. A _r\ 
linear veloc. B r 2 ' 

This is true whether the rotation is about a center or a centro ; 
i.e., it is true either for continuous or instantaneous rotation. 
Hence it applies to all cases of plane motion in machines; because 
all plane motion in machines is equivalent to either continuous 
or instantaneous rotation about some point. 

To find the relation of linear velocity of two points in a machine 
member, therefore, it is only necessary to find the relation of 



MOTION IN MECHANISMS. 25 

the radii of the points. The latter relation can easily be found 
when the center or centro is located. 

20. A vector quantity possesses magnitude and direction. It 
may be represented by a straight line, because the latter has 
magnitude (its length) and direction. Thus the length of a 
straight line, AB, may represent, upon some scale, the magnitude 
of some vector quantity, and it may represent the vector quantity's 
direction by being parallel to it, or by being perpendicular to it. 
For convenience the latter plan will here be used. The vector 
quantities to be represented are the linear velocities of points 
in mechanisms. The lines which represent vector quantities are 
called vectors. 

A line which represents the linear velocity of a point will 
be called the linear velocity vector of the point. The symbol of 
linear velocity will be VI. Thus VIA is the linear velocity of 
the point A. Also Va will be used as the symbol of angular 
velocity. 

If the linear velocity and radius of a point are known, the 
angular velocity, or the number of revolutions per unit time, 
may be found; since the linear velocity -f- length of the circum- 
ference in which the point travels = angular velocity. 

All points of a rigid body have the same angular velocity. 

If the radii, and ratio of linear velocities of two points, in 
different machine members are known, the ratio of the angular 
velocities of the members may be found as follows: 

Let A be a point in a member M, and B a point in a member 

N. r\ — radius of A ; r 2 = radius of B. VIA and VIB represent 

VIA 
the linear velocities of A and B, whose ratio, 777^, is known. 

V In 

VIA VIB 
Then VaA = and VaB= . 

2TZY\ 277T2 

VaA_VlA 2nr2_VlA r 2 VaM 
Hence VaB ~ 27zn X VIB ~ VIB X n ~ VaN ' 



26 



MACHINE DESIGN. 



If M and N rotate uniformly about fixed centers, the ratio 

VaM 

is constant. If either M or N rotates about a centro, the 



VaN 

ratio is a varying one. 

21. To find the relation of linear velocity of two points in 
the same link, it is only necessary to measure the radii of the 
points, and the ratio of these radii is the ratio of the linear veloci- 
ties of the points. 

In Fig. 1 6, let the smaller circle represent the path of A, 
the center of the crank-pin of a slider-crank mechanism; the 
link d being fixed. Let the larger circle represent the rim of a 
pulley which is keyed to the same shaft as the crank. The 
pulley and the crank are then parts of the same link. The ratio 

VIA 



of velocity of the crank-pin center and the pulley surface =- 



VIB 



= — . In this case the link rotates about a fixed center. The 

same relation holds, however, when the link rotates about a 
centro. 




Fig. i 6. 

22. Velocity Diagram of Slider-crank Chain. — In Fig. 17, 

Vlab ab-bd 
the link d is fixed and 7777—=^ — rr- By similar triangles this 
Vlbc bc-bd J b 

expression is also equal to n _. . Hence, if the radius of the 



MOTION IN MECHANISMS. 



27 



crank circle be taken as the vector of the constant linear velocity 
of ab, the distance cut off on the vertical through O by the line of 
the connecting-rod (extended if necessary) will be the vector of the 
linear velocity of be. Project A horizontally upon bc-bd, locating 
B. Then bc-B is the vector of VI of the slider, and may be 

Fig. 17. 




used as an ordinate of the linear velocity diagram of the slider. 
By repeating the above construction for a series of positions, 
the ordinates representing the VI of be for different positions of 
the slider may be found. A smooth curve through the extremi- 
ties of these ordinates is the velocity curve, from which the Vis 



28 MACHINE DESIGN. 

of all points of the slider's stroke may be read. The scale of 
velocities, or the linear velocity represented by one inch of ordi- 
nate, equals the constant linear velocity of ab divided by O-ab 
in inches. 

23. Velocity Diagram of Lever-crank Chain. — It is required 
to find VI of be during a cycle of action of the mechanism shown 
in Fig. 18, d being fixed, and VI of ab being constant. The 
two points ab and be may both be considered in the link b. 
All points in b move about bd relatively to the fixed link. 

Vlab ab-bd 

Hence nfri^bl- 

For most positions of the mechanism bd will be so located as to 
make it practically impossible to measure these radii, but a line, 
as MN, drawn parallel to b cuts off on the radii portions which 
are proportional to the radii themselves, and hence proportional 
to the Vis of the points. Hence 

Vlab ab-M 



Vlbc bc-N ' 

The arc in which be moves may be divided into any number of 
parts, and the corresponding positions of ab may be located. A 
circle through M, with ad as center, may be drawn, and the 
constant radial distance ab-M may represent the constant 
velocity of ab. Through Mi, M2, etc., draw lines parallel to the 
corresponding positions of b, and these lines will cut off on the 
corresponding line of c a distance which represents VI of be. 
Through the points thus determined the velocity diagram may 
be drawn, and the VI of be for a complete cycle is determined. 
The scale of velocities is found as in Sec. 22. 

24. The relation of linear velocity of points not in the same 
link may also be found. 



MOTION IN MECHANISMS. 



29 



Required T// . R referred to d as the fixed link, Fig. 19. 

The centro ab is a point in common to a and b, the two links 
considered. Consider ab as a point in a; and its VI is to that 
of A as their radii or distances from ad. Draw a vector triangle 
with its sides parallel to the triangle formed by joining A, ab> 




Fig. 19. 

and ad. Then if the side A\ represent the VI of A, the side ai&i 
will represent the VI of ab. Consider ab as a point in b, and 
its VI is to that of B as their radii, or distances to bd. Upon 
the vector aib\ draw a triangle whose sides are parallel to those 
of a triangle formed by joining ab, bd, and B. Then, from 
similar triangles, the side Bi is the vector of j3's linear velocity. 



Hence 



VI of A vector A 1 
VI of B = vector B{ 



The path of B during a complete cycle may be traced, and the 
VI for a series of points may be found, by the above method ; then 
the vectors may be laid off on normals to the path through the 
points; the velocity curve may be drawn; and the velocity of 
B at all points becomes known. 



3o 



MACHINE DESIGN. 



In general, to find the instantaneous motion, in direction and 
velocity, of a point X in any link x, relatively to a fixed link d, 
given the motion of any other link a relatively to d: First, locate 
the centro ad. All points of a rotate about this centro relatively 
to d and have linear velocities proportional to their distances 
from ad. Second, locate centro ax. Its velocity relatively to 
d is known, likewise its direction of motion, since it is a point 
of a and therefore rotates about ad with a linear velocity pro- 
portional to its distance from ad. As a point of x it has this 
same motion relatively to d. Third, locate dx. All points of 
x relatively to d rotate about this centro and have linear velocities 
proportional to their distances from dx. Hence the linear veloc- 
ity of the point X is to the known linear velocity of the point 

X— dx 

flias — . 

ax — dx 

25. Angularity of Connecting-rod. — The diagram of VI of 
the slider-crank mechanism, Fig. 17, is unsymmetrical with 
respect to a vertical axis through its center. This is due to the 
angularity of the connecting-rod, and may be explained as follows : 
In Fig. 20, AO is one angular position of the crank, and BO 
is the corresponding angular position on the other side of the 

vertical through the 
center of rotation. 
The corresponding 
positions of the slider 
are as shown. But 
for position A the 
line of the connecting- 
rod, C, cuts off on 
the vertical through 
O a vector Oa, which represents the slider's velocity. For posi- 
tion B the vector of the slider's velocity is Ob and the velocity 
diagram is unsymmetrical. 




Fig. 20. 



MOTION IN MECHANISMS. 3 1 

If the connecting-rod were parallel to the direction of the 
slider's motion in all positions, as in the slotted cross head 
mechanism (see Fig. 14), the vector cut off on the vertical through 
O would be the same for position A and position B and the 
velocity diagram would be symmetrical. 

Since the velocity diagram is symmetrical with a parallel 
connecting-rod and unsymmetrical with an angular connecting- 
rod, with all other conditions constant, it follows that the lack 
of symmetry is due to the angularity of the connecting-rod. 

The velocity diagram for the slotted cross-head mechanism 
is symmetrical with respect to both vertical and horizontal axes 
through its center. In fact, if the crank radius (= length of 
link a) be taken as the vector of the VI of ab, the linear velocity 
diagram of the slider becomes a circle whose radius = the length of 
the link a. Hence the crank circle itself serves for the linear velocity 
diagram, the horizontal diameter representing the path of the slider. 

26. Angularity of Connecting-rod, Continued. — During a por- 
tion of the cycle of the slider-crank mechanism, the slider's VI 
is greater than that of ab. 
This is also due to the an- 
gularity of the connecting-rod, 
and may be explained as fol- 
lows: In Fig. 21, as the crank 

n. y 

moves up from the position 

1 r Fig. 21. 

x, it will reach such a position, 

A, that the line of the connecting-rod extended will pass 
through B. OB in this position is the vector of the linear veloc- 
ity of both ab and the slider, and hence their linear velocities 
are equal. When ab reaches B, the line of the connecting-rod 
passes through B ; and again the vectors — and hence the linear 
velocities — of ab and the slider are equal. For all positions 
between A and B the line of the connecting-rod will cut OB 
outside of the crank circle ; and hence the linear velocity of the 



ab ^ab 




32 MACHINE DESIGN. 

slider will be greater than that of ab. This result is due to the 
angularity of the connecting-rod, because if the latter remained 
always horizontal, its line could never cut OB outside the circle. 
It follows that in the slotted cross-head mechanism the maximum 
VI of the slider = the constant VI of ab. The angular space BOA, 
Fig. 21, throughout which VI of the slider is greater than the VI 
of ab, increases with increase of angularity of the connecting-rod; 
i.e., it increases with the ratio 

Length of crank 
Length of connecting-rod * 

27. Quick-return Mechanisms. — A slider in a mechanism 
often carries a cutting-tool, which cuts during its motion in one 
direction, and is idle during the return stroke. Sometimes the 
slider carries the piece to be cut, and the cutting occurs while 
it passes under a tool made fast to the fixed link, the return 
stroke being idle. 

The velocity of cutting is limited. If the limiting velocity 
be exceeded, the tool becomes so hot that it becomes unfit for 
cutting. The limit of cutting velocity depends on the nature 
of the material to be cut, and the quality of the tool-steel used. 
There is no limit of this kind, however, to the velocity during 
the idle stroke; and it is desirable to make it as great as possible, 
in order to increase the product of the machine. This leads 
to the design and use of " quick-return" mechanisms. 

28. Slider-crank Quick Return. — If, in a slider-crank 
mechanism, the center of rotation of the crank be moved, so 
that the line of the slider's motion does not pass through it, the 
slider will have a quick-return motion. 

In Fig. 22, when the slider is in its extreme position at the 
right, A, the crank-pin center is at D. When the slider is at B, 
the crank-pin center is at C. If rotation is as indicated by the 
arrow, then, while the slider moves from B to A, the crank-pin 




Fig. 2: 



MOTION IN MECHANISMS. 33 

center moves from C over to D. And while the slider returns from 
A to B, the crank-pin center moves under from D to C. If the 
VI of the crank-pin center be as- 
sumed constant, the time occu- 
pied in moving from D to C is 
less than that from C to D. 
Hence the time occupied by the 
slider in moving from B to A is 
greater than that occupied in 
moving from A to B. The mean 
velocity during the forward 
stroke is therefore less than during the return stroke. Or the 
slider has a " quick-return " motion. 

It is required to design a mechanism of this kind for a length 
of stroke =BA and for a ratio 

mean VI forward stroke 5 
mean VI return stroke 7 * 

The mean velocity of either stroke is inversely proportional to 
the time occupied, and the time is proportional to the correspond- 
ing angle described by the crank. Hence 

mean velocity forward 5 angle /? 
mean velocity return 7 angle a 

It is therefore necessary to divide 360 into two parts which 
are to each other as 5 to 7. Hence a =210° and /? = i5o°. Ob- 
viously = i8o° — /? = 30°. Place the 30 angle of a drawing 
triangle so that its sides pass through B and A. This condition 
may be fulfilled and yet the vertex of the triangle may occupy 
an indefinite number of positions. By trial O may be located so 
that the crank shall not interfere with the line of the slider.* 



*To avoid cramping of the mechanism, the angle BAD should equal or exceed 
i35°- 



34 MACHINE DESIGN. 

O being located tentatively, it is necessary to find the correspond- 
ing lengths of crank a and connecting-rod b. When the crank- 
pin center is at D, AO = b -a; when it is at C, BO = b -fa. AO 
and BO are measurable values of length whose difference = 2a ; 
hence a and b may be found, the crank circle may be drawn, 
and the velocity diagrams may be constructed as in Fig. 17; 
remembering that the distance cut off upon a vertical through O, 
by the line of the connecting-rod, is the vector of the VI of the 
slider for the corresponding position when the VI of the crank- 
pin center is represented by the crank radius. 

It is required to make the maximum velocity of the forward 
stroke of the slider = 20 feet per minute, and to find the corre- 
sponding number of revolutions per minute of the crank. The 
maximum linear velocity vector of the forward stroke = the 
maximum height of the upper part of the velocity diagram* 
call it Vl\. Call the linear velocity vector of the crank-pin center 
F/2= crank radius. Let x = linear velocity of the crank-pin 

center. Then 

Vl\ 20 ft. per minute 

vT 2 = x ' 

20 ft. per minute X VI2 

x- — Wi . 

x is therefore expressed in known terms. If now x, the space 
the crank-pin center is required to move through per minute, 
be divided by the space moved through per revolution, the result 
will equal the number of revolutions per minute =N; 

N 



27rX length of crank* 



29. Lever-crank Quick Return — Fig. 23 shows a compound 
mechanism. The link d is the supporting frame or fixed link, 
and a rotates about ad in the direction indicated, communicating 



MOTION IN MECHANISMS. 



35 



motion to c through the slider b so that c vibrates about cd. The 
link e, connected to c by a turning pair at ce, causes / to slide 
horizontally on another part of the frame or fixed link d. The 
center of the crank-pin, ab, is given a constant linear velocity, 
and the slider, /, has motion toward the left with a certain mean 
velocity, and returns toward the right with a greater mean velocity. 
This is true because the slider / moves toward the left while a 
moves through the angle a ; and toward the right while a moves 
through the angle /?. But the motion of a is uniform, and hence 
the angular movement a represents more time than the angular 
movement /?; and /, therefore, has more time to move toward 
the left than it has to move through the same space toward the 
right. It therefore has a " quick-return " motion. 



nd# 




Fig. 23. 

The machine is driven so that the crank-pin center moves 
uniformly, and the velocity, at all points of its stroke, of the 
slider carrying a cutting-tool, is required. The problem, there- 
fore, is to find the relation of linear velocities of ef and ab for a 
series of positions during the cycle; and to draw the diagram 
of velocity of ef. 

Solution. — ab has a constant known linear velocity. The 
point in the link c which coincides, for the instant, with ab, re- 
ceives motion from ab, but the direction of its motion is different 



36 MACHINE DESIGN. 

from that of ab, because ab rotates about ad, while the coin- 
ciding point of c rotates about cd. If ab-A be laid off repre- 
senting the linear velocity of ab, then ab-B will represent the 
linear velocity of the coinciding point of the link c. Let the 
latter point be called x. 

Locate cj, at the intersection of e with the line cd-ad. Now 
cj and x are both points in the link c, and hence their linear 
velocities, relatively to the fixed link d, are proportional to their 
distances from cd. These two distances may be measured 
directly, and with the known value of linear velocity of x = ab-B 
give three known values of a simple proportion, from which the 
fourth term, the linear velocity of cj, may be found. 

Or, if the line BD be drawn parallel to cd-ad, the triangle 
B-D-ab is similar to the triangle cd-cj-ab, and from the simi- 
larity of these triangles it follows that BD represents the linear 
velocity of cj on the same scale that ab-B represents the linear 
velocity of x. Hence the linear velocity of cj, for the assumed 
position of the mechanism, becomes known. But since cj is a 
point of the slider, all of whose points have the same linear velocity 
because its motion relatively to d is rectilinear translation, it 
follows that the linear velocity of cj is the required linear velocity 
of the slider. At ef erect a line perpendicular to the direction 
of motion of the slider having a length equal to BD. 

This solution may be made for as many positions of the 
mechanism as are necessary to locate accurately the velocity 
curve. The ordinates of this curve will, of course, be the veloci- 
ties of the slider, and the abscissas the corresponding positions 
of the slider. 

Having drawn the velocity diagram, suppose that it is required 
to make the maximum linear velocity of the slider on the slow 
stroke =Q feet per minute. Then the linear velocity of the 
crank-pin center ab=y can be determined from the propor- 
tion 



MOTION IN MECHANISMS. 37 

y vector A-ab 

Q maximum ordinate of velocity diagram' 

_ vector A-ab 

J ^maximum ordinate of velocity diagram* 

y 

If r = the crank radius, the number of revolutions per minute = — . 

When this mechanism is embodied in a machine, a becomes 
a crank attached to a shaft whose axis is at ad. The shaft turns 
in bearings provided in the machine frame. The crank carries a 
pin whose axis is at ab } and this pin turns in a bearing in the 
sliding block b. The link c becomes a lever keyed to a shaft 
whose axis is at cd. This lever has a long slot in which the block 
b slides. The link e becomes a connecting-rod, connected to both 
c and / by pin and bearing. The link / becomes the ''cutter- 
bar" or "ram" of a shaper: the part which carries the cutting- 
tool. The link d becomes the frame of the machine, which not 
only affords support to the shafts at ad and cd, and the guiding 
surfaces for /, but also is so designed as to afford means for holding 
the pieces to be planed, and supports the feed mechanism. 




Fig. 24. 



30. Whitworth Quick Return. — Fig. 24 shows another com- 
pound linkage, d is fixed, and c rotates uniformly about cd, 



38 MACHINE DESIGN. 

communicating an irregular rotary motion to a through the slider 
b. a is extended past ad (the part extended being in another 
parallel plane), and moves a slider / through the medium of 
a link e. This is called the " Whitworth quick-return mechanism." 
The point be, at which c communicates motion to a, moves along 
a, and hence the radius (measured from ad) of the point at which 
a receives a constant linear velocity varies, and the angular 
velocity of a must vary inversely. Hence the angular velocity 
of a is a maximum when the radius is a minimum, i.e., when 
a and c are vertical downward; and the angular velocity of a 
is minimum when the radius is a maximum, i.e., when a and c 
are vertical upward. 

31. Problem. — To design a Whitworth Quick Return for a 

given ratio, 

mean VI of / forward 

mean VI of / returning' 

When the center of the crank-pin, C, reaches A, the point D will 

coincide with B, the link c will occupy the angular position cd-B, 

and the slider / will be at its extreme position toward the left. 

When the point C reaches F, the point D will coincide with 
E, the link c will occupy the angular position cd-E, and the 
slider / will be at its extreme position toward the right. 

Obviously, while the link c moves over from the position 
cd-E to the position cd-B, the slider / will complete its forward 
stroke, i.e., from right to left. While c moves under from cd-B 
to cd-E, / will complete the return stroke, i.e., from left to right. 
The link c moves with a uniform angular velocity, and hence the 
mean velocity of / forward is inversely proportional to the angle 
P (because the time consumed for the stroke is proportional to 
the angle moved through by the crank c), and the mean velocity 
of / returning is inversely proportional to a. Or 

mean VI of / forward a 
mean VI of / returning /?' 



MOTION IN MECHANISMS. 39 

For the design the distance cd-ad must be known. This may 
usually be decided on from the limiting sizes of the journals at cd 

and ad. Suppose that the above ratio =-5-=—, that cd-ad = 3", 

and that the maximum length of stroke of /=i2 // . Locate cd 
and measure off vertically downward a distance equal to 3", 
thus locating ad. Draw a horizontal line through ad. The 
point ej of the slider / will move along this line. Since 

a 5 

j=-, and a+/? = 36o°, 

,\ a = 150° and /? = 2io°. 

Lay off a from cd as a center, so that the vertical line through 
cd bisects it. Draw a circle through B with cd as a center, B 
being the point of intersection of the bounding line of a with a 
horizontal through ad. The length of the link c = cd-B. 

The radius ad-C must equal the travel of /-f-2=6". This 
radius is made adjustable, so that the length of stroke may be 
varied. The connecting-rod, e, may be made of any convenient 
length. 

32. Problem. — To draw the velocity diagram of the slider 
/ of the Whitworth Quick Return. The point be, Fig. 25, as a 
point of c has a known constant linear velocity relative to d, and 
its direction of motion is always at right angles to a line joining 
it to cd. That point of the link a which coincides in this posi- 
tion of the mechanism with be, receives motion from be, but its 
direction of motion relative to d is at right angles to the line be- 
ad. If be- A represents the linear velocity of be, its projection 
upon bc-ad extended will represent the linear velocity of the 
point of a which coincides with be. Call this point x. Locate 
the centro a], draw the line af-bc and extend it to meet the vertical 
dropped from B to C. The centro a] may be considered as a 



4° 



MACHINE DESIGN. 



point in a, and its linear velocity relative to d, when so considered, 
is proportional to its distance from ad. Hence 

VI of af _ ad-af 
VI of x ad-bc ' 




Fig. 25. 
But the triangles ad-aj-bc and B-C-bc are similar. Hence 

VI of af BC 
Vloix~ B-bc 

This means that BC represents the linear velocity of af upon the 
same scale that B-bc represents the linear velocity of x. But 
af is a point in /, and all points in / have the same linear velocity 
relative to d since the motion is rectilinear translation; hence 
BC represents the linear velocity of the slider / for the given 
position of the mechanism, and it may be laid off as an ordinate 
of the velocity curve. This solution may be made for as many 
positions as are required to locate accurately the entire velocity 
curve for a cycle of the mechanism. 



CHAPTER III. 

PARALLEL OR STRAIGHT-LINE MOTIONS. 

33. Watt Parallel Motion. — Rectilinear motion in machines 
is usually obtained by means of prismatic guides. It is some- 
times necessary, however, to accomplish the same result by 
linkages. 

The simplest and most widely known linkage used for giving 
rectilinear motion to a point without the use of any sliding pairs 
is the so-called Watt Parallel Motion. It is one of the numerous 
inventions of James Watt and N 

consists of four links, three • 2 \ a b 

moving and one fixed, all con- 
nected by turning pairs, d, 
Fig. 26, is the fixed link, a 

rotates relative to d about ad; 

Fig. 26. 
c rotates relative to a about ca. 

The mechanism is shown in the position corresponding to the 

middle of its motion. 

As the points ab and be swing in the dotted arcs, the point P 

will travel in approximately a straight line. The whole path of 

P is a lemniscate, but the part which is ordinarily used approaches 

very closely to a straight line. 

34. Parallelogram. — A true parallel motion is given by the 
Parallelogram which is shown in Fig. 27. The links a, b, e t and e 
are connected to each other by turning pairs, and the linkage is 
attached to the fixed link d by a turning pair at ac. (This point 
is also ad and cd.) 

41 




42 



MACHINE DESIGN. 



The lengths ac — ab and ce — be are equal, as are also ac — ce and 
ab — be. The point P is fixed on e. Draw a line from P to ac ; it 
cuts the link 6 at P'. 

By similar triangles, 

P'-fo P-6e 



ac — ce 



.". P' — be =ac — ce 



ce 
be 



[P-ce) 



a constant. 



Therefore the point P f will lie at the same position on b for all 
positions of the mechanism. Likewise, by similar triangles, the 

P —ac P — r e 

ratio — = — = a constant for all positions of the 

P' — ac be — ce 

mechanism. Since the line P — P' swings, relative to d, about the 




Fig. 27. Fig. 28. 

pole ad (ac, cd) at every instant, it is obvious that the motions of 
P and P' relative to d will be similar to each other in every respect 

P -ac 



and always in the ratio 



It follows that, if either of 



ac 



these points is guided to move in a straight-line path, the other 
point is constrained to move in a similar parallel path.* 

* The following demonstration is given for those who prefer an accurate 
proof : 

The position of the mechanism in Fig. 27 is taken as a perfectly general one 
and, in the same way, the instan'aneous motion of the point P is assumed as 
indicated by the arrow. By methods indicated in earlier chapters locate the 
centros de and bd. Continue the lines ad — ab and be — ab until they cut the line 
P — de at x and y, respectively. It is necessary to prove that P f will always 



PARALLEL OR STRAIGHT-LINE MOTIONS. 43 

35. Grasshopper Motion. — A device which may be used to 
change the direction of rectilinear motion through a right angle 
lis the linkage known as the Grasshopper Motion, shown in Fig. 28. 
This is the ordinary slider-crank chain with crank a and con- 
Inecting-rod b of equal length. The linkage is further modified 

I move in a path parallel to P's motion and at a constant proportion to it. If 
I this is true for instantaneous motion it is true for any motion P may be given 
relatively to d. Draw the line P' — bd. It can be shown that this line will 
always be parallel to P — de, for, since ac — x is parallel to ce — P 



Also, by similar triangles 
Hence, 



p- 


-be 


x- 


-bd 


be 


—ce 


bd 


— ac 


P- 


-be 


P- 


-P' 


be 


—ce 


P' 


—ac 


x- 


-Id 


P- 


-P' 



bd—ac P' —ac 7 



which, considering the triangles ac — x — P and ac — bd— P', shows that P' — bd 
is parallel to P— x, or to P — de. 

But P is a point of e and as such has an instantaneous motion relative to d in 
a direction perpendicular to P — de. In the same way, P' is a point of b and as 
such, relatively to d } has instantaneous motion perpendicular to P' — bd. These 
two instantaneous motions are therefore parallel. 

It remains to be shown that they will always be in the same proportion as to 

extent. The extent will be directly proportional to the instantaneous linear 

velocities. Both P and be as points of e rotate for the instant about de 

, . , J VIP P-de 

relatively to a, .". — — = 7 -. 

' Vibe b^-de 

Both be and P' are points of b, and as such, relatively to d, rotate about bd, 
Vibe b?-bd be-de , L . „ . . , 

•• YTP' = T^bd =J=Te (by Slmilar man s les )- 

Mu'tipl)ing, 

VIP Vibe P-de be-de VIP P-de „ . ., . , . 

?fi — X 77-777= r-X r—> or -xr, 1 / = r- = (by similar trianglee) , 

Vloe ViP' be—dey—de' VIP' < u — de w 6 " 

P-be 



be—ce 



a constant value. 



Or, in other words, the linear veloci.ies of P and P' bear a constant ratio to each 
other for all positions of the mechanism, and hence, these points will trace 
proportionately similar paths one/. — Q.E.D. 



44 



MACHINE DESIGN. 



by extending b beyond ab to a point P such that the length 
P-ab = crank length. It is obvious that P is constrained to move 
relative to d in a straight-line path perpendicular to d through 
ad* 

36. General Method for Parallel-motion Design. — A general 
method of design which is applicable in many cases is as follows. 
In Fig. 29 d is the fixed link, and a is connected with it by a sliding 
pair, a, b, c, and e are connected by turning pairs, as shown. 
The constrainment is not complete because B is free to move 
in any direction, and its motion would, therefore, depend upon 
the force producing it. It is required that the point B shall move 



ce be c 




Fig. 29. 



in a straight line parallel to a. Suppose that B is caused to move 
along the required line; then any point of the link c, as A, will 
describe some curve, FAE. If a pin be attached to c, with its 
axis at A, and a curved slot fitting the pin, with its sides parallel 
to FAE, be attached to d, as in Fig. 30, it follows that B can only 
move in the required straight line. This is the mechanism of the 
Tabor Steam-engine Indicator. 

* This is true because, from the construction of the mechanism, the line P-bd 
must always lie parallel to d. The point P, which rotates about bd as center rela- 
tively to d, always has an instantaneous motion perpendicular to P-bd and, con- 
sequently, perpendicular to d. 



PARALLEL OR STRAIGHT-LINE MOTIONS. 



45 



The curve described by A might approximate a circular arc 
whose center could be located, say, at O, Fig. 30. Then the 




Fig. 30. 

curved slot might be replaced by a link attached to d and c by 
turning pairs at O and A. This gives B approximately the 
required motion. This is the mechanism of the Thompson 
Steam-engine Indicator. 

If, while the point B is caused to move in the required straight 
line, a point in b, as P, Fig. 29, were chosen, it would be found 




Fig. 31. 

to describe a curve which would approximate a circular arc, 
whose center, O, and radius, =r, could be found. Let the link 



4 6 



MACHINE DESIGN 



whose length = r be attached to d and b by turning pairs whose 
axes are at O and P, and the motion of B will be approximately 
the required motion. This is the mechanism of the Crosby 
Steam-engine Indicator. One very important fact, however, 
is to be noted in connection with all steam-engine indicator 
pencil mechanisms. While it is important that the pencil point 




ce 2 aud ed a 



Fig. 30J . 

B (Figs. 29 and 30) travel in a straight-line path parallel to the 
axis of the piston rod a, it is fully as important that the motion 
of the point B always be exactly the same multiple of a's motion. 
To determine in any given case whether this is true or not, lay 
off very accurately and to a large scale, say five times actual size, 
a skeleton outline of the mechanism for three positions. See 
Fig. 30^4. These positions are taken so that the total distance 



PARALLEL OR STRAIGHT-LINE MOTIONS. 47 

B t — B 3 represents the allowable range of the instrument as 
stated by the maker, usually about 3". B 2 is located at the mid- 
position. The links a, b, c, e and /are drawn for each case in 
their proper relative positions, d being considered as the fixed 
link. 

The subscripts 1, 2 and 3 refer to the positions of the links 
corresponding to the three pencil positions B v B 2 and B s . 

First take position B v The centros df and de are located 
at once because they are permanent centers as well. Since a's 
motion relative to d is rectilinear translation, the centro ad will 
lie at infinity in a direction at a right angle to the direction of 
motion, or, in this case, at horizontal infinity. The centros cf v 
ab v bc v and ce 1 are located at once at the axes of the turning 
pairs connecting the respective links. 

Using Kennedy's theorem locate cd 1 (on lines de—ce 1 and 

df—cfj, and ac 1 (on lines cd 1 — ad and ab 1 — bc l ). At the 

instant in question, every point of c relatively to d is rotating 

about the centro cd x and each point will have a linear velocity 

proportional to its distance from cd v But B l and ac 1 are both 

. Viae, ac—cd ^ 

points of c. Hence we may write * = — - 1 • But ac t 

VIJ3 1 I> 1 — cd t 

is also a point of a and at any instant every point of a has the 

same velocity relatively to d because the motion of a relative to d 

is rectilinear translation .*. — — — = — J ■* • 

VIB X B x —edi 

Similarly for the second position, 

VI a _ ac 2 — cd 2 
VI B 2 = B 2 - cd 2 ' 

And for the third position, 

VI a ac a — cd 3 



VI B, B.-cd. 



3 ""-3 



48 MACHINE DESIGN. 



But, for proper action, 

Via Via Via 
VI B x ~ VI B 2 ~ VI B 3 



a constant for all positions, 



ac.—cd. , ,, , ac 9 — ca„ . . . ac., — cd v 

.*. — J - 1 should equal — J -^ and also equal — - f, 

B x — cd x B 2 —cd 2 , B 3 — cd 3 

otherwise the diagrams will give a distortion of the piston, a's, 
motion. Also for true parallel motion B x should lie on the same 
horizontal through cd x on which ac x lies; B 2 on the horizontal 
through cd 2 ) and B 3 on the horizontal through cd 3 . 

An examination of existing indicator mechanisms in this 
manner gives very interesting results, and separates clearly 
those instruments which distort from those which are correct. 

37. Problem. — In Fig. 31, B is the fixed axis of a counter- 
shaft; C is the axis of another shaft which is free to move in 
any direction. It is required to constrain D to move in the 
straight line EF. If D be moved along EF, a tracing-point 
fixed at A in the link CD will describe an approximate circular 
arc, HAK, whose center may be found at O. A link whose 
length is OA may be connected to the fixed link, and to the link 
CD, by means of turning pairs at O and A. D will then be 
constrained to move approximately along EF. A curved slot 
and pin could be used, and the motion would be exact. 

37A. Peaucellier Straight-line Motion. — This is an eight- 
link chain, all link connections being turning pairs, which gives 
true straight-line motion within the limits of its action. The 
mechanism is shown in two positions in Fig. 31^4, one in heavy 
lines, the other in light lines. The fixed link is d, and when a 
is oscillated about the centro ad, the point P travels in a straight 
line. The linkage is symmetrical: a = d, b = c, and e, f, g, and 
h are equal. Because of this symmetry, whatever position of 
the mechanism be considered, A and P will lie in a straight line 
passing through D, as will also the mid-point O of the diagonal 
A-P. Since e, /, g, and h are equal, it can be proven readily 



PARALLEL OR STRAIGHT-LINE MOTIONS. 



49 



that the diagonal B-C bisects A-P at O and makes an angle 
of 90 with it. 



Therefore, 



and 



DC 2 = DO 2 + CO 2 , 

cp 2 ==cd 2 +po 2 , 

DC 2 -CP 2 = D0 2 -P0 2 , 



= (DO-PO)(DO+PO), 
= DAxDP. 

DC and CP, being the length of the respective links c and f f 
are constant for the mechanism irrespective of position; from 




Fig. 31 A. 



which it follows that the first term in the foregoing equation 
is a constant and, therefore, the product of the variable distances 
DA and DP is a constant. 

Consider a to have swung to the position ad-A' '. The 
mechanism will then be constrained to assume the position shown 
by the light lines. From what has just been proven, 

DAXDP = DA'XDP', 

DA DP' 
*'* DA~DP' 



50 MACHINE DESIGN. 

The triangles DPP' and DA' A, having an angle in common 
with its adjacent sides proportional, are similar. But the end 
of link a in swinging from position A to A' travels in the cir- 
cumference of a circle whose diameter = DA (d and a being equal), 
and consequently the angle DA' A is a right angle. From this 
it follows that the angle DPP' must also be a right angle'. The 
proof, being independent of the position chosen, holds for all 
positions and, therefore, the point P travels in a straight line 
perpendicular to DP. 

The theoretical limits to this motion are the positions on 
either side of the line DP when DP' = (c+f)* 

* Descriptions of many varieties of parallel motions maybe found in Rankine's 
"Machinery and Millwork"; Weisbach's "Mechanics of Engineering," Vol. Ill, 
"Mechanics of the Machinery of Transmission"; Kennedy's "Mechanics of 
Machinery"; and elsewhere. 



CHAPTER IV. 

CAMS. 

38. Cams Defined. — A cam is a machine part of irregular 
outline which, having a motion of reciprocation, rotation or 
oscillation, communicates motion by means of point or line 
contact to another part, called the follower. In most cases the 
follower is to be given a motion of reciprocation or oscillation 
according to stated conditions governing the positions it is to 
occupy for definite positions of the cam. 

The method of design is the same in all cases, and consists 
in laying down in the plane of the cam the successive positions 
which the follower surface is to occupy relative to it, and then 
in drawing the outline of the cam tangent to these positions. 
It follows from this that the form of the cam outline depends 
not only on the law of motion which the follower is to obey, 
but also depends upon the form of the contact surface of the 
follower. These principles can be understood most readily by 
considering a few simple cases. 

39. Case 1. — A cam — which rotates counterclockwise, with 
uniform angular velocity, about a given center — is to impart 
a motion of straight-line reciprocation to a follower, the center 
line of whose action passes through the cam center. The fol- 
lower is to rise with uniform linear velocity from its lowest to 
its highest position while the cam rotates through 135 ; it is to 
remain at rest while the cam rotates through the next 45 ; and 
it is to return with uniform velocity to its lowest position while 
the cam completes its rotation. Three cams will be considered: 
(a) the follower is pointed, i.e., wedge-shaped; (b) the follower 

51 



52 



MACHINE DESIGN. 




Fig. 32. 



has a cylindrical roller; and (c) the follower has a flat contact 
surface normal to its line of action. 

Consider first the case with the pointed follower shown in 
Fig. 32. With a center at O, the center of rotation of the cam, 

a circle is drawn through the point 
occupied by the tip of the follower in 
its lowest position. This is known as 
the base circle. This circle is now 
divided, in a clockwise direction 
(against the direction of cam rotation), 
into three major portions, of 135 , 45 , 
and 180 , respectively, corresponding 
to the major divisions of the follower's 
cycle of motion. The path of the 
follower point is next divided into 
any convenient number of parts, say 
three, equal in this case since the follower is to rise uniformly. 
The more divisions there are taken, the more accurately will 
the cam outline be determined. The 135 portion of the base 
circle is divided into the same number of equal parts by radial 
lines. Beginning with the lowest, number the follower point 
positions 1, 2, 3, and 4, on the right-hand side of the line of 
action. Beginning with the vertical radial line of the cam, num- 
ber these elements similarly 1, 2, 3, and 4. When element 1 of 
the cam is in the vertical position the follower tip is to be at the 
distance O-i from the cam center. When element 2 of the cam 
has swung up to the vertical position, the follower point is to be 
at the distance 0-2 from the cam center. With this distance 
as radius, and O as center, swing an arc cutting the cam element 
2 and at this point of intersection draw the follower tip, showing 
in the plane of the cam the simultaneous relation of cam and 
follower when element 2 reaches the vertical position. Repeat 
this construction for elements 3 and 4. Since the follower remains 



CAMS. 



53 



stationary for the next 45 of the cam's rotation, the relative 
positions of cam element and follower surface are the same for 
element 5 as for 4. Divide the remaining 180 of the cam into 
as many equal parts as may be convenient, say 4, numbering 
the successive radial lines 5, 6, 7, 8, and r. Similarly divide the 
distance the follower is to descend during this angle of rotation 
of the cam into 4 equal parts, numbering the respective posi- 
tions of the follower point 5, 6, 7, 8, and 1, on the left-hand side 
of the line of action. (If the follower were not to descend with 
uniform velocity, these divisions would, of course, no longer be 
made equal, but would be laid off in direct proportion to the 
motion to be imparted to it.) Continue the layout of simul- 
taneous positions for this portion of the cam. Draw a smooth 
curve through the positions occupied 
by the follower points on the radial 
lines — the portion from elements 4 
to 5 being a circular arc. This 
gives the outline of the cam which 
will give the desired motion to the 
follower. 

Example (b) is constructed in 
exactly the same manner. See Fig. 
33. In this case the base circle is 
drawn through the lowest position 
of the follower roller center. The 
positions which this center is to 

occupy for the various angles of cam rotation are located, just 
as were the pointed follower tips, and swung to the respective 
radial lines. At each center thus located a circle, the size of the 
follower roller, is drawn and the cam outline completed by 
drawing a smooth curve tangent to these circles. 

Example (c), Fig. 34, shows the identical construction applied 
to the flat-faced follower — the successive simultaneous positions 




Fig. 33. 



54 



MACHINE DESIGN. 



of cam element and lower surface of follower being laid down 
and the cam outline drawn by constructing a smooth tangent 
curve to the follower surfaces. This type of follower frequently 

calls for impossible cams, unless the 
given law of motion maybe mod- 
ified slightly. 

It will be noted that, although 
the law of motion for the follower 
is the same in all three cases, the 
form of the cam outline derived 
varies with the form of contact sur- 
face of the follower. 

In the cases considered above 
the follower is given a uniform mo- 
tion up and down. This is done for 
simplicity. As a usual rule, particularly for high-speed operation, 
the follower neither should be started nor stopped abruptly. 
It should not be given a uniform motion throughout its path, 
but should be given a gradually accelerated motion at the start 
in each direction and a similarly retarded motion at the finish. 




Fig. 34. 





Fig. 35. 



In Fig. 35 (a) there is shown a method, which explains itself, 
for so dividing the path, AB, of the follower that the latter will 
be given accelerations and retardations corresponding to simple 
harmonic motion. In Fig. 35 (b) there is shown a construction 
for conveniently dividing the follower path AB to give still more 



CAMS. 



55 



gradual starting and stopping accelerations and retardations. 
The law is that of gravity acceleration. Both cases show sym- 
metrical motion of the follower about the mid-point, C, of its 
path. This is, obviously, not essential. Fig. 35 (c) shows the 
necessary modification to adapt method (a) for making the fol- 
lower have its maximum velocity at any point, D, of its path- 
The method is equally applicable to construction (b). 

40. Case II. — The follower is pointed and is to have a motion 
of straight-line reciprocation, but now its line of action passes 
to one side of, and not through, the center of rotation of the 
cam. The line may be in any position relatively to the center 
of rotation of the cam; hence it is a general case. The point 
of the follower which bears on the cam is constrained to move 
in the line MN, Fig. 36. O is the center of rotation of the cam. 
About O as a center, draw a circle 
tangent to MN at /. Then A, B, 
C, etc., are points in the cam, 
found by dividing the base circle 
with radial lines corresponding to 
the angles through which the cam 
is to rotate while the follower oc- 
cupies successive positions, accord- 
ing to the method described in the 
preceding section. When the point 
A is at / the point of the follower FlG - 3<5 ' 

which bears on the cam must be at A'; when B is at / the 
follower point must be at B' ; and so on through an entire revolu- 
tion. Through A, B, C, etc., draw lines tangent to the circle. 
With O as a center, and OA' as a radius, draw a circular arc 
A' A", intersecting the tangent through A at A" . Then A" 
will be a point in the cam curve. For, if A returns to /, A A" 
will coincide with J A ' , A" will coincide with A', and the cam 
will hold the follower in the required position. The same process 




56 



MACHINE DESIGN. 



for the other positions locates other points of the cam curve. 
A smooth curve drawn through these points is the required 
cam outline. Often, to reduce friction, a roller attached to 
the follower rests on the cam, motion being communicated 
through it. The curve found as above will be the path of the 
axis of the roller. The cam outline will then be a curve drawn 
inside of, and parallel to the path of the axis of the roller, at 
a distance from it equal to the roller's radius. The path of the 




Fig. 37. 



Fig. 37 A. 



point of contact between the follower and the cam is not confined 
to the line MN if a roller is used. 

41. Case III — This is the same as Case I (c), except that 
the positions of the follower surface, instead of being parallel, 
converge to a point, O', Fig. 37, about which the follower vibrates. 
The solution is the same as in Fig. 34, except that the angle 
between the lines representing the corresponding positions of the 
lower, or contact, surface of the follower, and the radial lines, 
instead of being a right angle, equals the angle between the 
corresponding positions of the follower surface and the vertical. 



CAMS. 



57 



In these cases the cam drives the follower in only one direc- 
tion; the force of gravity, the expansive force of a spring, or 
some other force must hold it in contact with the cam. To 
drive the follower in both directions, the cam surface must be 




Fig. 38. 



double, i.e., it takes the form of a groove engaging with a pin 
or roller attached to the follower, as in Fig. 37A 

This method is inclined to produce excessive wear. A better 
method is to have the follower provided with two rollers on 
opposite sides of the cam-shaft engaging complementary cams. 
See Fig. 38. 

Cam A is designed to give the desired motion to the follower 
through the medium of roller 1. Every position of this roller 



58 



MACHINE DESIGN. 



causes roller 2 to occupy a definite position, and the complementary 
cam B is so designed as to correspond to these positions of roller 
2. Cam B is rigidly mounted on the same shaft as A, so that 




Fig. 38/L 





Fig. 38C. 



Fig. 38B. 



the two cams have no motion relative to each other. If the line 
of action of the follower passes through the center of the cam. 
shaft as shown in Fig. 38, it becomes a very simple matter to 
draw the outline of cam B; all that is necessary is to keep the 



CAMS. 59 

sum of the radial lengths a + b = & constant = the distance be- 
tween the centers of rollers i and 2. 

This method is also applicable to cams engaging flat-faced 
followers. In this case the complementary cams operate on 
parallel faces of a follower yoke as shown in Fig. 38 A. 

In the special case in which the law of motion of the fol- 
lower for one -half the revolution of the cam may be the reverse 
of that for the other half- re volution, it is not necessary to use 
complementary cams. A follower with two rollers can then 
be used on a single cam as shown in Fig. 38Z?. One half the 
cam is first designed (see full lines) to give the follower the de- 
sired motion in one direction. The other half of the cam (see 
dotted lines) is then determined by the condition that the fol- 
lower rollers must always be the same distance apart, hence 
A'-B' = A-B, etc. 

The same construction can be applied to a forked or yoke 
follower, as shown in Fig. 38C, the distance between the parallel 
tangents being uniformly equal to A-B. 

42. Case IV. — To lay out a cam groove on the surface of a 
cylinder to give the follower a motion of reciprocation parallel 
to the cam axis. — A, Fig. 39, is a cylinder which is to rotate 
continuously about its axis. B can only move parallel to the 
axis of A. B may have a projecting roller to engage with a 
groove in the surface of A. CD is the axis of the roller in its 
mid-position. EF is the development of the surface of the cylinder. 
During the first quarter-revolution of A, CD is required to move 
one inch toward the right with a constant velocity. Lay off 
GH= 1", and HJ = jKF, locating /. Draw GJ, which will be the 
middle line of the cam-groove. During the next half-revolution of 
A the roller is required to move two inches toward the left with a 
uniformly accelerated velocity. Lay off JL= 2" , and LM = \KF . 
Divide LM into any number of equal parts, say four. Divide 
JL into four parts, so that each is greater than the preceding one 



6o 



MACHINE DESIGN. 



by an equal increment. This may be done as follows: 1+2 + 
3 + 4=10.* Lay of from /, 0.1JL, locating a; then 0.2 J L from 
a, locating b; and so on. Through a, b, and c draw vertical 
lines; through m, n, and draw horizontal lines. The in- 
intersections locate d, e, and/. Through these points draw the 
curve from / to M, which will be the required middle line 
of the cam-groove. During the remaining quarter-revolution 
the roller is required to return to its starting point with a 




Fig. 39. 



uniformly accelerated velocity. The curve MN is drawn in 
the same way as JM. Using the line GJMN as a locus of 
centers, swing in circles of a diameter equal to that of the fol- 
lower roller. The envelope of these is the developed cam groove. 
This groove can now be projected back, from the developed 
cylinder to the cylinder A, by the ordinary method of descriptive 

* The most frequent type of uniformly accelerated motion is that of falling 
bodies, in which the successive distances gone through in equal intervals of time 
are in the ratios: 1, 3, 5, 7, etc. 



CAMS. 



61 



geometry. In other words, wrap EF upon the cylinder A, and 
the required cam groove is located. 

The same method is applicable for determining a cam groove 
on a conical surface to give the follower a motion parallel to 
the cone surface element. 




The fundamental method laid down in this chapter is capable 
of indefinite application. Fig. 39^. shows a method by which 
two cams, operating simultaneously on two attached follower 
links, can cause a given point of the mechanism to trace almost 
any closed, plane path, however complex. Nor is the method 
confined to generating plane motion. The wide usefulness of 
cams in machines requiring intricate motions is, therefore, ap- 
parent. 



CHAPTER V. 

ENERGY IN MACHINES. 

43. The subject of motion and velocity, in certain simple 
machines, has been treated and illustrated. It remains now to 
consider the passage of energy through similar machines. From 
this the solution of force problems will follow. 

During the passage of energy through a machine, or chain of 
machines, any one, or all, of four changes may occur. 

I. The energy may be transferred in space. Example. — En- 
ergy is received at one end of a shaft and transferred to the other 
end, where it is received and utilized by a machine. 

II. The energy may be converted into another form. Exam- 
ples. — (a) Heat energy into mechanical energy by the steam- 
engine machine chain, (b) Mechanical energy into heat by fric- 
tion. U) Mechanical energy into electrical energy, as in a 
dynamo-electric machine; or electrical energy into mechanical 
energy in the electric motor, etc. 

III. Energy is the product of a force factor and a space factor. 
Energy per unit time, or rate of doing work, is the product of a 
force factor and a velocity factor, since velocity is space per unit 
time. Either factor may be changed at the expense of the other; 
i.e., velocity may be changed, if accompanied by such a change 
of force that the energy per unit time remains constant. Cor- 
respondingly, force may be changed at the expense of velocity, 
energy per unit time being constant. Example. — A belt trans- 
mits 6000 foot-pounds per minute to a machine. The belt veloc- 
ity is 120 feet per minute, and the force exerted is 50 lbs. Fric- 

62 



ENERGY IN MACHINES. 63 

tional resistance is neglected. A cutting-tool in the machine 
does useful work; its velocity is 20 feet per minute, and the re- 
sistance to cutting is 300 lbs. Then, energy received per minute 
= 120X50 = 60x30 foot-pounds; and energy delivered per minute 
= 20X300 = 6000 foot-pounds. The energy received therefore 
equals the energy delivered. But the velocity and force factors 
are quite different in the two cases. 

IV. Energy may be transferred in time. In many machines 
the energy received at every instant equals that delivered. There 
are many cases, however, where there is a periodical demand for 
work, i.e., a fluctuation in the rate of doing wcrk; while energy 
can only be supplied at the average rate. Or there may be a uni- 
form rate of doing work, and a fluctuating rate of supplying 
energy. In such cases means are provided in the machine, or 
chain of machines, for the storing 0) energy till it is needed. In 
other words, energy is transferred in time. Examples. — (a) In 
the steam-engine there is a varying rate of supplying energy dur- 
ing each stroke, while there is (in general) a uniform rate of doing 
work. There is, therefore, a periodical excess and deficiency of 
effort. A heavy wheel on the main shaft absorbs the excess of 
energy with increased velocity, and gives it out again with re- 
duced velocity when the effort is deficient, (b) A pump delivers 
water into a pipe system under pressure. The water is used in a 
hydraulic press, whose action is periodic and beyond the capacity 
of the pump. A hydraulic accumulator is attached to the pipe 
system, and while the press is idle the pump slowly raises the 
accumulator weight, thereby storing potential energy, which is 
given out rapidly by the descending weight for a short time while 
the press acts, (c) A dynamo-electric machine is run by a steam- 
engine, and the electrical energy is delivered and stored in storage 
batteries, upon which there is a periodical demand. In this case, 
as well as in case (b), there is a transformation of energy as well 
as a transfer in time. 



64 MACHINE DESIGN, 

44. Force Problems. — Suppose the slider-crank mechanism in 
Fig. 40 to represent a shaping-machine, the velocity diagram of 




Fig. 40. 

the slider being drawn. The resistance offered to cutting metal 
durinsr the forward stroke must be overcome. This resistance 
may be assumed constant. Throughout the cutting stroke there 
is a continually varying rate 0} doing work. This is because the 
rate of doing work = resisting force (constant) X velocity (vary- 
ing). This product is continually varying, and is a maximum 
when the slider's velocity is a maximum. The slider must be 
driven by means of energy transmitted through the crank a. The 
maximum rate at which energy must be supplied equals the maxi- 
mum rate of doing work at the slider. Draw the mechanism in 
the position of maximum velocity of slider; * i.e., locate the center 
of the slider-pin at the base of the maximum ordinate of the veloc- 
ity diagram, and draw b and a in their corresponding positions. 
The slider's known velocity is represented by y, and the crank- 
pin's required velocity is represented by a on the same scale. 
Hence the value of a becomes known by simple proportion. The 
rate of doing work must be the same at c and at ab (neglecting 
friction).f Hence Rv\= ify 2 , in which R and v\ represent the 

* It is customary to assume the slider's position for this condition to be that 
corresponding to an angle of 90 between crank and connecting-rod. This is not 
exactly true, but is a sufficiently close approximation for the ordinary proportions 
of crank and connecting-rod lengths. For method of exact determination of 
slider's position see Appendix. 

t The effect of acceleration to redistribute energy is zero in this position, be- 
cause the acceleration of the slider at maximum velocity is zero, and the angular 
acceleration of b can only produce pressure in the journal at ad. If R a equals 



ENERGY IN MACHINES. 6$ 

force and velocity factors at c; and F and v 2 represent the tangen- 
tial force and velocity factors ah. R and v t are known from the 
conditions of the problem, and v 2 is found as above. Hence F may 

Rvi 
be found, = = force which, applied tangentially to the crank- 

pin center, will overcome the maximum resistance of the machine. 
In all other positions of the cutting stroke the rate of doing work 
is less, and F would be less. But it is necessary to provide driv- 
ing mechanism capable of overcoming the maximum resistance, 
when no fly-wheel is used. If now F be multiplied by the crank 
radius, the product equals the maximum torsional moment ( = M) 
required to drive the machine. If the energy is received on some 
different radius, as in case of gear or belt transmission, the maxi- 
mum driving force = M -r- the new radius. During the return 
stroke the cutting-tool is idle, and it is only necessary to overcome 
the frictional resistance to motion of the bearing surfaces. Hence 
the return stroke is not considered in designing the driving mech- 
anism. When the method of driving this machine is decided on, 
the capacity of the driving mechanism must be such that it shall 
be capable of supplying to the crank-shaft the torsional driving 
moment M, determined as above. 

This method applies as. well to the quick-return mechanisms 
given. In each, when the velocity diagram is drawn, the vector 
of the maximum linear velocity of the slider, *=Zi, and of the 
constant linear velocity of the crank-pin center, = L 2 , are known, 
and the velocities corresponding, Vi and v 2 > are also known, from 
the scale of velocities. The rate of doing work at the slider and 

the force necessary to produce acceleration of the slider mass at any position and 
F a the force necessary at the crank pin to produce tangential acceleration of the 
rotating mass (assuming a variable velocity of the crank-pin as well as slider), 
then the equation in its most general form will be (R-\-Ba)vi = (F -\-F a )v<t. 
With uniform rate of rotation of the crank this becomes (R + R a )v 1 = Fv 2 ); and 
for position corresponding to maximum velocity of slider, as above, Rvi=Fv 2 . 



66 MACHINE DESIGN. 

at the crank-pin center is the same, friction being neglected. 
Hence Rv\ =Fv 2 , or, since the vector lengths are proportional 

RLi 

to the velocities they represent, RLi = FL 2 ; and F=-j — . There- 

fore the resistance to the slider's motion, =R, on the cutting 

stroke, multiplied by the ratio of linear velocity vectors, ■=-, of 

slider and crank-pin, equals F, the maximum force that must be 
applied tangentially at the crank-pin center to insure motion. 
F multiplied by the crank radius = maximum torsional driving 
moment required by the crank-shaft. If R is varying and known, 
find where Rv, the rate of doing work, is a maximum, and solve 
for that position in the same way as above. 

Where the mass to be accelerated is considerable the maxi- 
mum effort will be called for at the beginning of each stroke. 
If there is a quick return the maximum effort will come at the 
beginning of the return stroke. A planer calls for about twice 
as much power at the beginning of its return stroke as it does 
during its cutting stroke. 

45. Force Problems, Continued. — In the usual type of steam- 
engine the slider-crank mechanism is used, but energy is supplied 
to the slider (which represents piston, piston-rod, and cross- 
head), and the resistance opposes the rotation of the crank and 
attached shaft. In any position of the mechanism (Fig. 41), 
force applied to the crank-pin through the connecting-rod may 
be resolved into two components, one radial and one tangential. 
The tangential component tends to produce rotation; the radial 
component produces pressure between the surfaces of the shaft- 
journal and its bearing. The tangential component is approxi- 
mately a maximum when the angle between crank and connecting- 
rod equals 90 ,* and it becomes zero when C reaches A or B. 
If there is a uniform resistance the rate of doing work is constant. 

* See foot-note on page 64. 



ENERGY IN MACHINES. 67 

Hence, since the energy is supplied at a varying rate, it follows 
that during part of the revolution the effort is greater than the 
resistance; while during the remaining portion of the revolution 
the effort is less than the resistance, and the machine will stop 
unless other means are provided to maintain motion. A "fly- 
wheel " is keyed to the shaft, and this wheel, because of slight 




Fig. 41. 

variations of velocity, alternately stores and gives out the excess 
and deficiency of energy of the effort, thereby adapting it to the 
constant work to be done.* 

46. Problem. — Given length of stroke of the slider of a steam- 
engine slider-crank mechanism, the required horse-power, 01 
rate of doing work, and number of revolutions. Required the 
total mean pressure that must be applied to the piston. 

Let L = length of stroke = 1 foot ; 
HP = horse-power = 20; 
N = strokes per minute =200; 
F = required mean force on piston. 
Then N X L = 200 feet per minute = mean velocity of slider = V. 

Now, the mean rate of doing work in the cylinder and at the 
main shaft during each stroke is the same (friction neglected); 
hence FV=HPx 33000, 

HP X 33000 20X33000 

F= f7 = =3300 lbs. 

V 200 °° 

*See Chapter XVI. 



68 



MACHINE DESIGN. 



47. Solution of Force Problem in the Slider-crank Chain. 

—In the slider-crank chain the velocity of the slider necessarily 
varies from zero at the ends of its stroke to a maximum value 
near mid-stroke. The mass of the slider and attached parts 
is therefore positively and negatively accelerated each stroke. 
When a mass is positively accelerated it stores energy; and 
when it is negatively accelerated it gives out energy. The amount 
of this energy, stored or given out, depends upon the mass and 
the acceleration. The slider stores energy during the first part 
of its stroke and gives it out during the second part of its stroke. 




^HL 



Fig. 42. 

While, therefore, it gives out all the energy it receives, it gives 
it out differently distributed. In order to find exactly how the 
energy is distributed, it is necessary to find the acceleration 
throughout the slider's stroke. This may be done as follows: 
Fig. 42, A, shows the velocity diagram of the slider of a slider- 
crank mechanism for the forward stroke, the ordinates repre- 
senting velocities, the corresponding abscissae representing the 

Av 

Tv 



slider positions. The acceleration required at any point = 



in which Av is the increase in velocity during any interval of 



ENERGY IN MACHINES. 69 

time At, assuming that the increase in velocity becomes constant 
at that point. Lay off the horizontal line OP=MN. Divide 
OP into as many equal parts as there are unequal parts in MN. 
These divisions may each represent At. At m erect the ordinate 
mn = niini, and at erect the ordinate op = Oipi. Continue 
this construction throughout OP, and draw a curve through the 
upper extremities of the ordinates. Fig. 42, B, is a velocity 
diagram on a "time base." At O draw the tangent OT to the 
curve. If the increase in velocity were uniform during the time 
interval represented by Om, the increment of velocity would be 
represented by mT. Therefore mT is proportional to the accel- 
eration at the point O, and may be laid off as an ordinate of an 
acceleration diagram (Fig. 42C). Thus Qa = mT. The divi- 
sions of QR are the same as those of MN; i.e., they represent 
positions of the slider. This construction may be repeated for 
the other divisions of the curve B. Thus at n the tangent nT\ 
and horizontal nq are drawn, and g7\ is proportional to the 
acceleration at n, and is laid off as an ordinate be of the ac- 
celeration diagram. To find the value in acceleration units 
of Qa, mT is read off in velocity units =Av by the scale 
of ordinates of the velocity diagram. This value is divided 
by At, the time increment corresponding to Om. The result 

. . . Av 
of this division — - = acceleration at M in acceleration units. 
At 

At = the time of one stroke, or of one half revolution of the crank, 

divided by the number of divisions in OP. If the linear velocity 

of the center of the crank-pin in feet per second, =v, be repre- 

MN 
sented by the length of the crank radius = = a, then the scale 

of velocities, or velocity in feet per second for 1 inch of ordinate, 

v r.DN 

=— = — 7 — • D is the actual diameter of the crank circle, N 

a aoo ' 

is the number of revolutions per minute, and a is the crank radius 
measured on the figure. 



70 MACHINE DESIGN. 

The determination of the acceleration curve, by means of 
tangents drawn to the " time-base" velocity curve, has a serious 
drawback. The tangent lines are laid down by inspection, and slight 
inaccuracy in their location and construction may lead to consid- 
erable errors in the ordinates obtained for the acceleration curve. 

The following method is therefore suggested as an alternative. 

If one point is rotating about another point with a given 

instantaneous velocity =v and a radius = r, the instantaneous 

v 2 
radial acceleration of either point toward the other = —. 

Consider the slider-crank chain in the position at the begin- 
ning of the forward stroke as shown in Fig. 43^ . The problem 
is to determine the acceleration of the point be toward ad. Ac- 
celerations toward the right will be considered as positive, toward 
the left as negative. In the position chosen the point ab is mov- 
ing, relatively to both links d and c, in the direction of the arrow 
with a velocity =v, the uniform velocity of ab relatively to d. 

The acceleration of be toward ad is always made up of two 

components, namely, the acceleration of be toward ab and the 

acceleration of ab toward ad. In the position under considera- 

v 2 
tion the acceleration of be toward ^=~r in a positive direction. 

v 2 
Similarly the acceleration of ab toward ad=— in a positive di- 
rection. The total acceleration of be toward ad therefore equals 

. _ . v 2 v 2 

the sum of these two components, or = T" + ~- 

On the other hand, at the end of the forward stroke, shown 

v 2 
in Fig. 43.B, the acceleration of be toward ab = -r in a positive 

v 2 
direction as before, while the acceleration of ab toward ad=— 

a 

in a negative direction. The algebraic sum of these two com- 



ENERGY IN MACHINES. 



71 



v v 2 
ponents therefore = -r — — . This quantity will always have a 

negative value, since in the slider-crank mechanism a must al- 
ways be smaller than b. 

To construct the acceleration curve, lay off a length MN 







t y^ 


c 




ab. 




fc 


ad 


9 


p 




9 
k a — i\ 


i<. - 








Fig. 44. 



(Fig- 43Q proportionate to the length of the stroke of the slider. 



v 2 v 2 



\t M erect an ordinate, MP, whose value equals t- + — . It 

a 

is best to use for these ordinates a scale on which a (the length 



72 



MACHINE DESIGN. 



of the crank) represents the value — . At N erect the negative 



ordinate NQ = 



t.* 

a 



There is a position of the slider, O, where the acceleration 
equals zero. This must correspond to the position of the slider 

* The following construction for graphically obtaining ordinates representing 

v 2 . v 2 , v 2 v 2 , . ... 

■ > on the scale upon which a represents v 



v 2 , v 2 
— + — and - 
b a b 



( and, hence, a = — j is due to Professor Le Conte. 




Fig. 44A. 

Reference is to Fig. 44^4. M and N represent the position of the slider at the 
beginning and end of the stroke, respectively. 

At ab x erect the vector v ( = a) and from M draw a line through its upper extremity. 
Prolong this line until it cuts the perpendicular through ad, thus determining the 
length y v 

At ab 2 lay off downward the vector v (=a). From N draw a line to the lower 
extremity of this vector, cutting off the length y 2 on the perpendicular tlirough ad 



Then will y x represent— H — ; and y 2 represent - — 
For, taking slider position if, by similar triangles, 



y, r _ v(=a) 



a + b 



a 3 + ab 



+ a. But a 



Jx 



v 2 , v 2 



For position N, by similar triangles, 

a 2 — ab v 2 
.'. y *=—b— = b- C 



M- = lLi^l 



a—b 






ENERGY IN MACHINES. 73 

when it has its maximum velocity, which may be taken from 
the original velocity diagram of the slider, or, with greater 
accuracy, from Curve B in the Appendix. Through POQ draw 
a smooth curve. For most purposes this curve will be accurate 
enough. 

Where more points of the curve are desired for the sake of 
greater accuracy the method illustrated in Fig. 44 may be 
used. Assume the slider in the position at which its acceleration 
is desired and draw the crank a and connecting-rod b in their 
corresponding positions. Locate the centros ad, ab, ac, and bd. 
From ac draw a parallel to bc-ad until it cuts the crank, pro- 
longed if necessary, at A. From A draw a parallel to ad-ac 
until it cuts the connecting-rod at B. From B draw a perpen- 
dicular to the connecting-rod until it cuts bc-ad, prolonged if 
necessary, at C. Then ad-C is the desired ordinate of the 

v 2 
acceleration diagram on the scale by which the length a=— . 

The proof is as follows, reference being made to Fig. 44. 

At this instant every point of b relatively to c is swinging 
about the centro be with a velocity proportional to its distance 
from be. 

vel. of bd rel. to c bd-bc ac-ad 
vel. of ab rel. to c ab-bc ac-ab' 

But ac-ad represents the velocity of c relatively to d (or d 
relatively to c) on the same scale that ad-ab represents the 
velocity of the point ab relatively to d. Therefore ac-ab repre- 
sents the velocity of ab rotating about be relatively to c on the same 
scale that ad-ab represents the velocity of ab relatively to d. 

Hence the radial acceleration of ab toward be * (or conversely 

* The acceleration of a point A with respect to another point B is the accel- 
eration of A with respect to a non-rotating body of which B is a point. 



74 MACHINE DESIGN. 



ab—ac 
of be toward ab) = — 7 — , which is represented by the length 

ab-B, as can be shown as follows: 
By similar triangles 

ab-B ab-A ab-ac 
ab-ac ab-ad ab-bc' 



ah-B ob-ac ab-ac 



ab-bc b 

ab—aa 

The radial acceleration of ab toward ad= , whose 

a 

value we represent by the length a. The component of this 

acceleration in the direction bc-ab=ab-D. 

The acceleration of be, relatively to d, along the path bc-ab 
is made up of two components: 1st, the acceleration of be toward 
ab(=B-ab) plus, 2d, the acceleration of ab relatively to d along 
the same path (=ab-D). 

In the position shown this algebraic sum is the negative 
quantity represented by B-D. But the actual direction of fc's 
acceleration relatively to d is along the line bc-ad. Its accelera- 
tion in this direction must therefore be the quantity whose com- 
ponent along ab-bc is B-D, namely, C-ad. q.e.d. 

If the weight W of parts accelerated is known, the force F 
necessary to produce the acceleration at any slider position may 
be found from the fundamental formula of mechanics, 

p being the acceleration corresponding to the position considered. 
If the ordinates of the acceleration diagram are taken as repre- 
senting the jorces which produce the acceleration, the diagram 
will have force ordinates and space abscissae, and areas will 
represent work. Thus, Qas, Fig. 42C, represents the work stored 



ENERGY IN MACHINES. 75 

during acceleration, and Rsd represents the work given out during 
retardation. Let MN, Fig. 45, represent the length of the 
slider's stroke and NC the resistance of cutting (uniform) on the 
same force scale as that by which Qa, Fig. 42 C, represents the 

Wp 
force — - at the beginning of the stroke ; then energy to do cutting 

o 

per stroke is represented by the area MBCN. But during the 
early part of the stroke the reciprocating parts must be acceler- 
ated, and the force necessary at the beginning, found as above, 
= BD=Qa. The driving-gear must, therefore, be able to over- 
come resistance equal to MB + BD. The acceleration, and hence 
the accelerating force, decreases as the slider advances, becoming 
zero at E. From E on the acceleration becomes negative, and 
hence the slider gives out energy and helps to overcome the resist- 
ance, and the driving-gear has only to furnish energy represented 
D by the area AEFN, though the work really 

B — ^^^C^ - ic done against resistance equals that repre- 

^ A sen ted by the area CEFN. The energy 
f represented by the difference of these areas, 

Fig. 45. =ACE, is that which is stored in the 

slider's mass during acceleration. Since by the law of con- 
servation of energy, energy given out per cycle = that received, 
it follows that area ^C£=area DEB, and area BCMN ' = 
ADMN. This redistribution of energy would seem to modify 
the problem on page 62, since that problem is based on the 
assumption of uniform resistance during cutting stroke. The 
position of maximum velocity of slider, however, corresponds to 
acceleration =0. The maximum rate of doing work, and the 
corresponding torsional driving moment at the crank-shaft would 
probably correspond to the same position, and would not be 
materially changed. In such machines as shapers, the accelera- 
tion and weight of slider are so small that the redistribution of 
energy is unimportant. 



7 6 



MACHINE DESIGN. 



48. Solution of the Force Problem in the Steam-engine Slider- 
crank Mechanism. (Slider represents piston with its rod, and the 
cross-head.) — The steam acts upon the piston with a pressure 
which varies during the stroke. The pressure is redistributed 
before reaching the cross-head pin, because the reciprocating parts 
are accelerated in the first part of the stroke, with accompanying 
storing of energy and reduction of pressure on the cross-head 
pin; and retarded in the second part of the stroke, with accom- 
panying giving out of energy and increase of pressure on the 
cross-head pin. Let the ordinates of the full line diagram above 
OX, Fig. 46.4, represent the total effective pressure on the piston 
throughout a stroke. Fig. 46B is the velocity diagram of slider. 





Cr^ 



^^ 




Fig. 46. 

Find the acceleration throughout stroke, and from this and the 
known value of weight of slider find the force due to acceleration. 
Draw diagram Fig. 46C, whose ordinates represent the force 
due to acceleration, upon the same force scale used in A. Lay 
off this diagram on OX as a base line, thereby locating the dotted 
line. The vertical ordinates between this dotted line and the 
upper line of A represent the pressure applied to the cross-head 
pin. These ordinates may be laid off from a horizontal base line, 
giving D. The product of the values of the corresponding ordinates 
of B and £>=the rate oj doing work throughout the stroke. Thus 
the value of GH in pounds X value of EF in feet per second =the 



ENERGY JN MACHINES. 77 

rate of doing work in foot-pounds per second upon the cross- 
head pin, when the center of the cross-head pin is at E. The 
rate of doing work at the crank-pin is the same as at the cross- 
head pin. Hence dividing this rate of doing work, = EFxGH, 
by the constant tangential velocity of the crank-pin center, gives 
the force acting tangentially on the crank-pin to produce rotation. 
The tangential forces acting throughout a half revolution of 
the crank may be thus found, and plotted upon a horizontal 
base line= length of half the crank circle (Fig. 47.8). The work 
done upon the piston, cross-head pin, and crank during a piston 
stroke is the same. Hence the areas of A and D, Fig. 46, are 
equal to each other, and to the area of the diagram, Fig. 47$. 
The forces acting along ,the connecting-rod for all positions 
during the piston stroke may be found by drawing force triangles 
with one side horizontal, one vertical, and one parallel to position 
of connecting-rod axis, the horizontal side being equal to the 
corresponding ordinate of Fig. 46D. The vertical sides of these 
triangles will represent the guide reaction, while the side parallel 
to the connecting-rod axis represents the force transmitted by 
the connecting-rod. 

The tangential forces acting on the crank-pin may be found 
graphically by the method shown in Fig. 47^ . Let GH repre- 
sent the net effective force acting in a horizontal direction at the 
center of the cross-head pin. 

It has been shown that EF represents the velocity of the 

slider on the same scale that EA represents that of the center of 

the crank-pin; also that the rate of doing work, after having 

made the necessary corrections for acceleration, is the same at 

the center of the crank-pin as at the slider, i.e., GHXEF =ta,n- 

gential force at center of crank-pin XEA. Hence the tangential 

EF 
force at center of crank-pin = GH X tt, . 



78 



MACHINE DESIGN. 



Lay off AB=GH, and draw BC parallel to EF. Then, by 
similar triangles, 

the tangential force acting at the crank-pin center for the assumed 
position of the mechanism, on the same scale as GH = net effective 
horizontal force on slider. 

Lay oRAD=BC. 

Following through this construction for a number of positions 
of the mechanisms, a polar diagram is determined which shows 




Fig. 47B. 

very clearly the relation existing between the varying tangential 
forces and the corresponding crank positions. Before this dia- 
gram may be used in the solution of the fly-wheel problem (see 
Chapter XVI) it should be transferred to a straight-line base 
whose length for one stroke equals the semi-circumference of the 
crank-pin circle. That is, the abscissae will be the distance moved 
through by the center of the crank-pin and the ordinates will be 
the corresponding radial intercepts AD. The diagram so ob- 
tained will be identical with that shown m Fig. 47.6. 



ENERGY IN MACHINES. 



79 



48A. General Method for Determining Velocity and Accele- 
ration Diagrams. — The construction used first in Sec. 47 for 
determining the acceleration diagram of the slider-crank chain 
is capable of wide adaptation. It can be used to determine not 
only the acceleration diagram, but also the velocity diagram (upon 

A a i 1 1 I I U6 



1 2 



B c 






Fig. 47C. 

which the acceleration diagram is based) for any point in any 
body, provided only that the path of the point be known together 
with the positions it occupies at the end of equal intervals of 
time throughout its cycle. See Fig. 47C. At A is shown the 
path, a-b, the point traverses in the given time. Starting at 
1, the positions 2, 3, 4, 5, 6, and 7 are those reached after sue- 



80 MACHINE DESIGN. 

cessive equal intervals of time. At B a convenient distance, 
c-d, is taken as base-line to represent the time required by the 
point to travel the distance a-b. c-d is divided into as many- 
equal parts as a-b has unequal parts, and these distances 1-2, 
2-3, etc., represent equal increments of time, At. At 2 on c-d 
erect the ordinate MN = distance 1-2 of a-b. Similarly at 3 on 
c-d, erect an ordinate = distance 1-3, of a-b. Continue for all 
points on c-d. The ordinate at 7 will, of course, equal a-b 
itself. Draw a smooth curve through the ends of these ordinates. 
This will give a displacement curve on a time base. The ordinates 
represent distances, the abscissae, time. At N draw the tan- 
gent NP and the horizontal NO. Then will OP represent the 

velocitv of the point when it is at position 2 ; for velocity = — = — , 

dt At 

and, since OP equals the displacement increment for the time 

As OP 

NO, if the velocity at AT" held for the entire interval NO, — = t^. 
J At NO 

OP expressed in distance units divided by NO in time units, 

gives the actual velocity of the point at position 2 in velocity 

units. It is clear since the same vector, =NO, is taken each 

time to represent At, that the intercepts OP, TU, etc., may be 

taken themselves as the velocity vectors for the respective 

positions. 

At C, then, which is the velocity diagram of the point on a 
time base, OP — which is the velocity vector for position 2 — is 
laid off as an ordinate at 2. Similarly for the other positions, 
as TU at 4. The smooth curve cPUd is then drawn. 

From the velocity diagram on the time base, since accelera- 

Av 
tion = — , the vectors are derived for the acceleration diagram 

as described in sec. 47. Such an acceleration diagram is shown 
at D, laid off on a time base. It could be transferred readily 
to the position base shown at A . 



CHAPTER VI. 

PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 

49. The size and form of machine parts * are governed by 
six main considerations: 

(1) The size and nature of the work to be accommodated (as 
the swing of engine-lathes, etc.). 

(2) The stresses which they have to endure. 

(3) The maintaining of truth and accuracy against wear, in- 
cluding all questions of lubrication. 

(4) The cost of production. 

(5) Appearance. 

(6) Properties of materials to be used. 

The first is a given condition in any problem ; the second will 
be discussed here; the third will be treated in the chapters on 
Journals and Sliding Surfaces; the fourth is touched upon here; 
the principles governing the fifth are treated in Chapter XIX 
and here. 

It is assumed in this and following chapters that the reader is 
familiar with the properties of the materials employed in machine 
construction, f and with the general principles of the science of 
mechanics. A few tables are appended to this chapter for 
convenience. 

50. The stresses acting on machine parts may be constant, 
variable, or suddenly applied. 



* On this general subject see an excellent article by Prof. Sweet in the Jour- 
nal of the Franklin Institute, 3d Series, Vol. 125, pp. 278-300. The reader is 
also referred to "A Manual of Machine Construction," by Mr. John Richards, 
and to the Introduction of this volume. 

t See Smith's "Materials of Machines." 

81 



82 MACHINE DESIGN. 

A constant stress is frequently spoken of as a steady, or 

DEAD, LOAD. 

A variable stress is known as a live load. 

A suddenly applied stress is known as a shock. 

51. Constant Stress. — If a machine part is subjected to a con- 
stant stress, i.e., an unvarying load constantly applied, its design 
becomes a simple matter, as the amount of such a stress can 
generally be very closely estimated. Knowing this and the 
properties of the materials to be used, it is only necessary to cal- 
culate the area which will sustain the load without excessive 
deformation. 

Thus, in simple tension or compression, if we let U = the ulti- 
mate strength of the material in pounds per square inch, F the 
total constant stress in pounds, A the unknown area in square 
inches necessary to sustain F, we write 



U+K' 



where K is a so-called factor of safety, introduced to reduce 
the permitted unit stress to such a point as will limit the deforma- 
tion (strain) to an allowable amount, and also to provide for pos- 
sible defects in the material itself. In exceptional cases where 
the stresses permit of accurate calculation, and the material is of 
proven high grade and positively known strength, K has been 
given as low a value as ij; but values of 2 and 3 are ordinarily 
used for wrought iron and steel free from welds; while 4 to 5 are 
as small as should be used for cast iron, on account of the uncer- 
tainty of its composition, the danger of sponginess of structure, 
and indeterminate shrinkage stresses. 



U . 
52. Variable Stress. — We pass next to the consideration of 



The safe unit stress =/=-~ in pounds per square inch 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 8$ 

variable stresses or live loads. Here the problem is much more 
complex than with dead loads. 

Experiments by Wohler,* and Bauschinger,f with the work 
of Weyrauch J and others have given us the laws of bodies sub- 
jected to repeated stresses. In substance Wohler's law is as 
follows: Material may be broken by repeated applications 

OF A FORCE WHICH WOULD BE INSUFFICIENT TO PRODUCE RUPTURE 
BY A SINGLE APPLICATION. THE BREAKING IS A FUNCTION OF 
RANGE OF STRESS; AND AS THE VALUE OF THE RECURRING STRESS 
INCREASES, THE RANGE NECESSARY TO PRODUCE RUPTURE DE- 
CREASES. If the stress be reversed, the range equals 

THE SUM OF THE POSITIVE AND NEGATIVE STRESS. 

Bauschinger's conclusions were as follows: 

(i) With repeated tensile stresses whose lower limit 

WAS ZERO, AND WHOSE UPPER LIMIT WAS NEAR THE ORIGINAL 
ELASTIC LIMIT, RUPTURE DID NOT OCCUR WITH FROM 5 TO 1 6 

million repetitions. He cautions the designer (a) that this 
will not hold for defective material, i.e., a factor of safety must 
still be used for this reason; and (b) that the elastic limit of the 
material must be carefully determined, because it may have been 
artificially raised by cold working, in which case it does not accur- 
ately represent the material. The original elastic limit may be de- 
termined by testing a piece of the material after careful annealing. 
(2) With often-repeated stresses varying between zero 
and an upper stress which is in the neighborhood of or 
above the elastic limit, the latter is raised even above, 
often far above, the upper limit of stress, and it is raised 
higher as the number of repetitions of stress increases, 

* "Ueber die Festigkeitsversuche mit Eisen und Stahl," A. Wohler, Berlin, 1870. 

f "Mittheilungen der Konig. Tech. Hochschule zu Miinchen," J. Bau- 
schinger, Munich, 1886 and 1897. 

X "Structures of Iron and Steel," by J. Weyrauch. Trans, by A„ J. DuBois, 
New York, 1877. 



84 MACHINE DESIGN. 

WITHOUT, HOWEVER, A KNOWN LIMITING VALUE, Z,, BEING EX- 
CEEDED. 

(3) Repeated stresses between zero and an upper limit 
below l do not cause rupture; but if the upper limit is 
above l rupture will occur after a limiting number of 
repetitions. 

From this it will be seen that keeping within the original 
elastic limit insures safety against rupture from repeated 
stress if the stress is not reversed; and that, when the stress is 
reversed, the total range should not exceed the original elas- 
tic range of the material. 

Various formulas have been proposed by different authorities 
embodying the foregoing laws. 

Unwin's is here given as being most simple and general: 

Let U be the breaking strength of the material in pounds 
per square inch for a load once gradually applied. 

Let / max. be the breaking strength in pounds per square inch 
for the same material subjected to a variable load ranging be- 
tween the limits /max. and /min., and repeated an indefinitely 
great number of times, /min. is + if the stress is of the same 
kind as /max., and is — if the stress is of the opposite kind, 
and it is supposed that / min. is not greater than / max. Then 
the range of stress is J=/max. T/min., the upper sign being 
taken if the stresses are of the same kind and the lower if they 
are different. Hence A is always positive. The formula * is 



/max.= — WU 2 —qAU, 



where y is a variable coefficient whose value has been experi- 
mentally determined. For ductile iron and steel 9 = 1.5, in- 
creasing with hardness of the material to a value of 2. 

* Unwin's "Machine Design," Vol. 1, 1903, pp. 32-36. 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 85 

This formula is of general application. 
Three cases may be considered: 

(1) A constant stress, or dead load. In this case the range 
of stress, A =0, and consequently / max. = U, as it should be. 

(2) The stress is variable between an upper limit and zero, 
but is not reversed. 

Here A = / max., since / min. =0, and consequently / max. = 

2(Vlj 2 + I-T))U. 

(3) The stress is reversed, being alternately a compressive 
and tensile stress of the same magnitude. 

Here / min. = — / max. and A = 2/ max. 



.*. /max. = — U. 

2T) 



In each case it is necessary to divide the breaking load, /max., 

by a factor of safety in order to get the safe unit stress /, i.e., 

f max 
/ = ' — — — -. K is a factor of safety whose numerical value depends 

upon the material used. (See Sec. 51.) 

53. Problem. — Consider that there are three pieces to be 
designed using machinery steel having an ultimate tensile strength 
of 60,000 lbs. per square inch. 

The first piece sustains a steady load having a dead weight 
suspended from it. 

The second piece is a member of a structure which is alter- 
nately loaded and unloaded without shock. 

The third piece is subjected to alternate stresses without 
shock. 

In each case the maximum load is the same, being 30,000 lbs. 
= F. This material is generally reliable and uniform in quality. 
A factor of safety of 3 is common; .*. K = $ i n eacn case; 9 = 1.5. 



86 MACHINE DESIGN. 

Case I. 

t TT A t / maX ' 

/ max. = U and / = 



U 60000 

.'. )=— = = 20,000 lbs. per sq. in. 

o o 

The necessary area A to sustain F is determined by the equation 

. F 30000 

A =~y = = it so. in. 

j 20000 z 

Case II. 

/max. =2(VV + i — 7?)Z7. 

17 = 1.5 and 27 = 60,000 lbs. per sq. in. 
.*. / max. =.6054^7 = 36,324 lbs. per sq. in. 

. / max. 

f = = 12,108 lbs. per sq. in. 

. F 30000 . 

7 = 12108 = 2 * ^- in -' ««wfy- 



Case III. 



/ max. = — £7. 

' 27? 



, 60OOO 

/ max. = = 20,000 lbs. per sq. in, 



/max. 

/ = =6667 lbs. per sq. in. 

o 

. J 7 30000 

The importance of considering the question of range of stress 
in designing is brought home by this illustration. Comparison 



PROPORTIONS OF MACHINE. PARTS AS DICTATED BY STRESS. 87 

of results shows that with the same maximum load in each case, 
the second piece must be given nearly twice the area of the first, 
while the third must be three times as great in area as the first, 
the only difference in the three cases being the range of stress. 

Table A (page 88), of allowable working stresses, as compiled 
by Unwin, is introduced for convenience of reference. Such 
tables are to be used with judgment in reference to the particular 
conditions in each case. 

54. Shock. — Consideration of the design of parts subjected 
to shocks or suddenly applied loads. 

(1) A load is applied on an unstrained member in a single 
instant, but without velocity. 

In this case, if the stress does not exceed the limit of elasticity 
of the material, the stress produced will be just twice that pro- 
duced by a gradually applied load of the same magnitude. If 
F = maximum total load as before, then the maximum total 
stress =2^. The design of the member is then made as in Case 
II or Case III of the preceding section. 

(2) A load is applied on an unstrained member in a single 
instant, but with velocity. 

In this case the stress on the member will exceed that due to 
a gradually applied load of the same magnitude by an amount 
depending on the energy possessed by the load at the moment 
of impingement. 

Assume that a member is stressed by a load F falling through 
a height h. The unknown area of the member =A, and the 
allowable strain (i.e., extension or compression) = L As before, 
/ = allowable unit stress (determined by the question of range by 
the use of Unwin 's formula). 

The energy of the falling load is F(h+ X). 

The work done in straining the member an amount X with a 

maximum fiber stress / is -XA, provided the elastic limit is 



88 



MACHINE DESIGN. 



Table A. — Ordinary Working Stress 

CASE A. DEAD LOAD INDUCING PERMANENT STRESS 



Material 




Kind of Stress 






Tension 


Compression 


Bending 


Shear 


Torsion 


Cast iron 

Wrought iron: 
Wrought bar or forged . 

Wrought plate 1 1 

Wrought plate j_ 


4200 

15000 

15000 

12000 

13000-17000 

1 7 000- 2 1 000 

8000-12000 

IOOOO 

4200 
4000 
3000 


12000 
15000 


6000- 8000 
15000 


2300 
12000 


4000- 6000 
7500 






IOOOO 

10000-13000 

13000-17000 

7000-12000 

7000 




13000- 1 7000 
17000-21000 
12000-16000 


13000-17000 
17000-21000 
10000-14000 


8000-12000 


Cast steel 


12000- 16000 

7000-12000 

4200 


Phosphor bronze 






Rolled copper 






2400 





















CASE B. VARYING LOAD. STRESS FROM ZERO TO A MAXIMUM 



Material 




Kind of Stress 




Tension 


Compression 


Bending 


Shear 


Torsion 




2800 

IOOOO 
IOOOO 

8000 

8600-1 1400 

11400-14000 

5300- 8000 

6600 

2800 

2600 

2000 


8000 

IOOOO 


4000- 5300 

IOOOO 


1500 
8000 


2600- 4000 
5000 


Bar iron 








6500 
6500- 8600 
8600-1 1400 
4700- 8000 

4600 




Mild steel 


8600-1 1400 

11400-14000 

8000-10600 


8600- 1 1400 

11400-14000 

6600- 9400 


5300- 8000 


Steel castings 

Phosphor bronze 


4700 -8000 
















1600 




Brass 



















CASE C. VARYING LOAD. ALTERNATE EQUAL STEESSES OF OPPOSITE SIGN 



Material 


Tension and 
Compression 


Bending and 
Bending 


Shear and 
Shear 


Torsion and 
Torsion 




1400 

5000 
4300-5700 
5700-7000 
2700-4000 

1400 


2000-2700 

5000 
4300-5700 
5700-7000 
3300-4700 


770 

4000 

3300-4300 

4300-5700 

2300-4000 






2500 
2700-4000 
4000-5300 
2300-4000 






Steel castings 











PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 89 

not exceeded. Equating these values of energy expended and 

2F(h + X) 
work done and solving for A gives A = 7+ . 

55. Form Dictated by Stress. Tension. — Suppose that A and 
B, Fig. 48, are two surfaces in a machine to be joined by a member 

subjected to simple tension. What is the proper 
form for the member? The stress in all sec- 
tions of the member at right angles to the line 
of application, ^4 Z2, of the force will be the same. 
Therefore the areas of all such sections should be equal; hence 
the outlines of the members should be straight lines parallel to 
AB. The distance of the material from the axis AB has no 
effect on its ability to resist tension. Therefore there is nothing 
in the character of the stress that indicates the form of the 
cross-section of the member. The form most cheaply produced, 
both in the rolling-mill and the machine-shop, is the cylindrical 
form. Economy, therefore, dictates the circular cross-section. 
After the required area necessary for safely resisting the stress is 
determined, it is only necessary to find the corresponding diam- 
eter, and it will be the diameter of all sections of the required 
member if they are made circular. Sometimes in order to get a 
more harmonious design, it is necessary to make the tension 
member just considered of rectangular cross-section, and this is 
allowable although it almost always costs more. The thin, wide 
rectangular section should be avoided, however, because of the 
difficulty of insuring a uniform distribution of stress. A unit 
stress might result from this at one edge greater than the strength 
of the material, and the piece would yield by tearing, although 
the average stress might not have exceeded a safe value. 

56. Compression. — If the stress be compression instead of 
tension, the same considerations dictate its form as long as it 
is a "short block," i.e., as long as the ratio of length to lateral 
dimensions is such that it is sure to yield by crushing instead of 



90 MACHINE DESIGN. 

by "buckling." A short block, therefore, should have its longi- 
tudinal outlines parallel to its axis, and its cross-section may be 
of any form that economy or appearance may dictate. Care 
should be taken, however, that the least lateral dimension of 
the member be not made so small that it is thereby converted 
into a "long column." 

If the ratio of longitudinal to lateral dimensions is such that 
the member becomes a "long column," the conditions that dic- 
tate the form are changed, because it would yield by buckling or 
flexure instead of crushing. The strength and stiffness of a 
long column are proportional to the moment of inertia of the 
cross-section about a gravity axis at right angles to the plane 
in which the flexure occurs. A long column with "fixed" or 
"rounded" ends has a tendency to yield by buckling which is 
equal in all directions. Therefore the moment of inertia needs 
to be the same about all gravity axes, and this of course points to 
a circular section. Also the moment of inertia should be as large 
as possible for a given weight of material, and this points to the 
hollow section. The disposition of the metal in a circular hollow 
section is the most economical one for long-column machine 
members with fixed or rounded ends. This form, like that for 
tension, may be changed to the rectangular hollow section if 
appearance requires such change. If the long-column machine 
member be "pin connected," the. tendency to buckle is greatest 
in a plane through the line of direction of the compressive force 
and at right angles to the axis of the pins. The moment of iner- 
tia of the cross-section should therefore be greatest about a gravity 
axis parallel to the axis of the pins. Example, a steam-engine 
connecting-rod. 

57. Flexure. — When the machine member is subjected to 
transverse stress the best form of cross-section is probably the I 
section, a, Fig. 49, in which a relatively large moment of inertia, 
with economy of material, is obtained by putting the excess of 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. Qi 

the material where it is most effective to resist flexure, i.e., at the 
greatest distance from the given gravity axis. Sometimes, how- 
ever, if the I section has to be produced by cutting away the 
material at e and d, in the machine-shop, instead of producing 
the form directly in the rolls, it is cheaper to use the solid rect- 
angular section c. If the member subjected to transverse stress 
is for any reason made of cast material, as is often the case, the 
form b is frequently preferable for the following reasons: 

(1) The best material is almost sure to be in the thinnest part 
of a casting, and therefore in this case at / and g, where it is most 
effective to resist flexure. 

(2) The pattern for the form b is more cheaply produced and 
maintained than that for a. The hollow box section, when 
permitted by considerations of construction and expense is still 
better. 

(3) If the surface is left without finishing from the mold, any 
imperfections due to the foundry work are more easily corrected 
in b than in a. 

Machine members subjected to transverse stress, which con-' 
tinually change their position relatively to the force which pro- 
duces the flexure, should have the same moment of inertia about 
all gravity axes, as, for instance, rotating shafts that are strained 
transversely by the force due to the weight of a fly-wheel, or that 
due to the tension of a driving-belt. The best form of cross- 
section in this case is circular. The hollow section would give 
the greatest economy of material, but hollow members are ex- 
pensive to produce in wrought material, such as is almost inva- 
riably used for shafts. The hollow circular section is meeting 
with increasing use, especially for large shafts, on account of the 
combined lightness and strength. 

58. Torsion. — Torsional strength and stiffness are propor- 
tional to the polar moment of inertia of the cross-section of the 
member. This is equal to the sum of the moments of inertia 
about two gravity axes at right angles to each other. The forms 



9 2 



MACHINE DESIGN. 



in Fig. 49 are therefore not correct forms for the resistance of 
torsion. The circular solid or hollow section, or the rect- 
angular solid or hollow section, should be used. 

The I section, Fig. 50, is a correct form for resisting the stress 
P, applied as shown. Suppose the web C to be divided on the 
line CD, and the parts to be moved out so that they occupy 
the positions shown at a and b. The form thus obtained is 
called a "box section." By making this change the moment 
of inertia about ab has not been changed, and therefore the 
new form is just as effective to resist flexure due to the force P 
as it was before the change. The box section is better able to 




VP 



C; 



flAft 



VqzltzlJ 



Fig. 50. 

resist torsional stress, because the change made to convert tne 
I section into the box section has increased the polar moment 
of inertia. The two forms are equally good to resist tensile and 
compressive force if they are sections of short blocks. But if 
they are both sections of long columns, the box section would be 
preferable, because the moments of inertia would be more nearly 
the same about all gravity axes. 

59. Machine Frames. — The framing of machines almost always 
sustains combined stresses, and if the combination of stresses 
include torsion, flexure in different planes, or long-column com- 
pression, the box section is the best form. In fact, the box sec- 
tion is by far the best form for the resisting of stress in machine 
frames. There are other reasons, too, besides the resisting of 
stress that favor its use.* 



* See Richard's "Manual of Machine Construction. 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. Q3 

(1) Its appearance is far finer, giving an idea of complete- 
ness that is always wanting in the ribbed frames. 

(2) The faces of a box frame are always available for the 
attachment of auxiliary parts without interfering with the per- 
fection of the design. 

(3) The strength can always be increased by decreasing the 
size of the core, without changing the external appearance of 
the frame, and therefore without any work whatever on the 
pattern itself. 

The cost of patterns for the two forms is probably not very 
different, the pattern itself being more expensive in the ribbed 
form, and the necessary core-boxes adding to the expense in the 
case of the box form. The expense of production in the foundry, 
however, is greater for the box form than for the ribbed form, 
because core work is more expensive than " green-sand " work. 
The balance of advantage is very greatly in favor of box forms, 
and this is now recognized in the practice of the best designers 
of machinery. 

To illustrate the application of the box form to machine 
members, let the table of a planer be considered. The cross- 
section is almost universally of the form shown in Fig. 51. This 
is evidently a form that would yield easily to a force tending to 



Fig. 51. 

twist it, or to a force acting in a vertical plane tending to bend it. 
Such forces may be brought upon it by "strapping down work," 
or by the support of heavy pieces upon centers. Thus in Fig. 52 
the heavy piece E is supported between the centers. For proper 
support the centers need to be screwed in with a considerable 
force. This causes a reaction tending to separate the centers and 
to bend the table between C and D. As a result of this the V's 



94 



MACHINE DESIGN. 



on the table no longer have a bearing throughout the entire 
surface of the guides on the bed, but only touch near the ends, 
the pressure is concentrated upon small surfaces, the lubricant is 
squeezed out, the V's and guides are "cut," and the planer is 




Fig. 52. 





Fig. 53. 



Fig. 54. 



rendered incapable of doing accurate work. If a table were made 
of the box form shown in Fig. 53, with partitions at intervals 
throughout its length, it would be far more capable of maintain- 
ing its accuracy of form under all kinds of stress, and would be 
more satisfactory for the purpose for which it is designed.* 

The bed of a planer is usually in the form shown in section 
in Fig. 54, the side members being connected by "cross-girts" 
at intervals. This is evidently not the best form to resist flexure 
and torsion, and a planer-bed may be subjected to both, either 
by reason of improper support or because of changes in the form 
of foundation. If the bed were of box section with cross parti- 
tions, it would sustain greater stress without undue yielding. 
Holes could be left in the top and bottom to admit of supporting 
the core in the mold, to serve for the removal of the core sand, 
and to render accessible the gearing and other mechanism inside 
of the bed. 



* Professor Sweet has designed and constructed such a table for a large mill- 
ing-machine. 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS 95 

This same reasoning applies to lathe-beds. They are strained 
transversely by force tending to separate the centers, as in the 
case of "chucking"; torsionally by the reaction of a tool cutting 
the surface of a piece of large diameter; and both torsion and 
flexure may result, as in the case of the planer-bed, from an 
improperly designed or yielding foundation. The box form 
would be the best possible form for a lathe-bed; some diffi- 
culties in adaptation, however, have prevented its extended use 
as yet. 

These examples illustrate principles that are of very broad 
application in the designing of machines. 

60. Brackets. — Often in machines there is a part that pro- 
jects either vertically or horizontally and sustains a transverse 





Fig. 55. 



stress; it is a cantilever, in fact. If only transverse stress is sus- 
tained, and the thickness is uniform, the outline for economy of 
material is parabolic. In such a case, however, the outline curve 
of the member should start from the point of application of the 
force, and not from the extreme end of the member, as in the 
latter case there would be an excess of material. Thus in A, 
Fig. 55, P is the extreme position at which the force can be ap- 



06 MACHINE DESIGN. 

plied. The parabolic curve a is drawn from the point of appli- 
cation of P. The end of the member is supported by the auxil- 
iary curve c. The curve b drawn from the end gives an excess 
of material. The curves a and c may be replaced by a single 
continuous curve as in C, or a tangent may be drawn to a at its 
middle point as in B, and this straight line used for the outline, 
the excess of material being slight in both cases. Most of the 
machine members of this kind, however, are subjected also to 
other stresses. Thus the "housings" of planers have to resist 
torsion and side flexure. They are very often supported by two 
members of parabolic outline; and to insure the resistance of the 
torsion and side flexure, these two members are connected at their 
parabolic edges by a web of metal that really converts them into 
a box form. Machine members of this kind may also be sup- 
ported by a brace, as in D. The brace is a compression member 
and maybe stiffened against buckling by a "web" as shown, or 
by an auxiliary brace. 

6 1. Other Considerations Governing Form. — One considera- 
tion governing the form of machine parts has been touched upon 
in the preceding sections. It may be well to state it here as a 
general principle: Other considerations being equal the form of 
a member should be that which can be most cheaply produced 
both as regards economy of material and labor. 

Another element enters into the form of cast members. Cast- 
ings, unless of the most simple form, are almost invariably sub- 
jected to indeterminate shrinkage stresses. Some of these are 
undoubtedly due to faulty work on the part of the molder, others 
are induced by the very form which is given the piece by the de- 
signer. They cannot be eliminated entirely, but the danger can 
be minimized by paying attention to these general laws: 

(a) Avoid all sharp corners and re-entrant angles. 

(b) All parts of all cross-sections of the member should be 
as nearly of the same thickness as possible. 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 97 

(c) If it is necessary to have thick and thin parts in the same 
casting, the change of form from one to the other should be as 
gradual as possible. 

(d) Castings should be made as thin as is consistent with con- 
siderations of strength, stiffness, and resistance to vibration. 

The following tables of properties of materials, and formulae 
of mechanics are grouped here for convenient reference. Tables 
of strength of materials mast be understood to represent ap- 
proximate average results. This property varies not only with 
the chemical constitution but also with the physical condition 
as affected by heat treatment, hot or cold- working, etc., and 
even by the size and form of the part. It is impossible to tabulate 
with reference to these factors. The machine designer must 
have an exhaustive knowledge of the properties of materials 
of construction and of the factors affecting these properties. 
Only such knowledge can guide him in the selection, say, of 
wrought iron rather than mild steel for a piece whose subsequent 
strength might be imperiled by overheating in the course of its 
manufacture; or, again, in the selection of a suitable tough, 
shock-resisting, alloy steel for a piece destined for severe service, 
which might readily fail if made of a harder, less yielding, but 
apparently " stronger " material. 



98 machine design. 

Table B. — Physical Properties of Metals. 



Metal 



Melting- 


Specific 




point 


gravity 


Struct- 


Degrees 


at ordin- 


ure* 


Fahr. 


ary temps 




1214 


2.65 


M 


Il66 


6.71 


B 


1562 


3-75 


M 


516 


9.80 


B 


6lO 


8.64 


M 


1436 


i-55 


M 


2712 


6.50 


B 


2714 


8.60 


M 


1982 


8.85 


M 


1944 


19.32 


M 


2 200f 


7.20 


B 


2975 


7.86 


M 


28oof 


7.70 


M 


621 


n-37 


S 


1172 


1.74 


M 


2205f 


8.00 


B 


-37-8 


13.60 


F 


2647 


8.80 


M 


3190 


21.50 


M 


144 


0.86 


S 


1760 


10.50 


M 


257°t 


7-7 


M 


257of 


7-85 


M 


449 


7.29 


M 


5544 


17.60 


B 


2947 


5-5o 


M 


784 


7. 10 


M 



Electric con- 
ductivity. 
Silver 100 at 
32 F. 



Approx. value 

per pound, 
Dollars (1915) 



Weight 

per cubic 

inch, 

Pounds 



Aluminum 

Antimony 

Barium 

Bismuth 

Cadmium 

Calcium 

Chromium 

Cobalt 

Copper 

Gold 

Iron, cast 

Iron, pure 

Iron, wrought. . . 

Lead 

Magnesium 

Manganese 

Mercury 

Nickel 

Platinum 

Potassium 

Silver 

Steel, machinery 

Steel, tool 

Tin 

Tungsten 

Vanadium 

Zinc 



50.2 
4.1 

1.1 

24.2 
21.3 

16.3 
92.2 
69.8 

1.9 
15.0 
14.8 

7.6 
35-2 

"i'o 

11. 7 

15-6 

6.4 

100. o 
10. o 

3-4t 

to 

I 10. 0% 

II .1 

33 -o 



27.0 



0.19 

0.15 

2800.00 

2.80 

i-35 
3.00 

i-5° 

4.00 

0.14 

300. oof 

O.OI 

0.02 
0.04 

2.50 
1-50 
0.7s 
0.50 
800 . 00 
16.00 

8.00 

0.02 
0.06 
to 
1. 00 

0.34 

i-75 

104.00 

0.06 



.0956 
.2421 
.1353 
3536 
.3118 

■0559 
.2346 

•3103 
•3194 
.6972 
.2598 
.2836 
.2779 
4103 
.0628 
.2887 
.4908 
•3175 
•7758 
•0314 
.3789 
.278 

•2833 

.2631 
.6351 
.1985 
.2562 



* B ^Brittle, F =Fluid, M = Malleable, S =Soft. 
t Varies. Approx. Av. Value Only, 
j Glass hard, 3.4. Soft, 10.00. 
§ $20.67 per ounce troy. 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 99 



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o 



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o o 

8 8 

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o~ o~ o~ 

OOO 
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o o 
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o o 

o~ o' 

o o 

o o 



H M MM CO 



& o 

OOO 
OOO 
°~ °~ °L 
vO~ 0~ ^r> 

CN CN 



ill M 



O O O O Q 

o o o o 

O O O O »o 





^^ 






el S 


a £ • 
!U : 




CU O 

+J 













8 : 
£ • 









IT) O 


^^0 • 




M O 


2^ : 




M M 


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C K 





en 






Tl 


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£ 


T3 





>> O 




<i> 


cfl 


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rri 





rfl 


O 

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Id 


Tl 


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rrt 


CJ 


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'p 


C 






P 


a 


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a^ 








a 


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cd 
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>> 


b-* 


0> 

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o o 

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co ^r 






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to <-» 

£ o 



CO 

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s 

o 

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8 tfT 

cu be 



w 



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BLISS'S 



0-2 _ v!-zi 



W Q£ Q OT 



cu 



c3 tn 



o 

^ cu 
P, So 

a, o 



§ 2 

be CS 

53 



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IOO 



MACHINE DESIGN. 



Table D. — Elements of Usual Sections. 



Moments refer to horizontal gravity axis, as shown. Values for flanged beams 
apply to standard minimum sections only. A =area of section. 



Shape of 
Section. 



Moment of 
Inertia, /. 



Section Modulus, 
/ 



Distance of 

Base from Center 

of Gravity. 



Least Radius 
of Gyration. 



*-b- 



12 



bh^ 

6 




0.1178& 3 



Least side 
3-46 



3-46 



—b, 



bS-b 2 * 



I &i 4 -6 2 4 

6 h 



< — b-, 




b¥_ 
36 



24 



V^? 



The lesser of 



4.24 4.9 




7T^ 
64 



ird* 
3 2 




o.o4Qi(di 4 — d 2 A ) 



0.0982 



Jl 4 -^2 



di 



iVdS+d 2 



£^ 



0.1098;- 4 



0.4244?* 



o.o699r 2 



irba 3 
~6T 



irba 2 
32 



<-B-^. 



PROPOR TIONS OF MA CHINE PAR TS AS DIC TA TED BY S TRESS. 1 01 



Table D. — Elements of Usual Sections. — Continued. 

Approximate. 



Shape of 
Section. 


Moment of 
Inertia, /. 


Section Modulus, 
/ 
c 


Distance of 

Base from Center 

of Gravity. 


Least Radius 
of Gyration. 




■i 


=,! 


Ah* 

10.4 


Ah 

7-4 


h 
3-5 


5 


< — ft — » 




^ 


y 


Ah 2 
6.1 


Aft 
3 


h 

2 


_b__ 
5-2 








Ah* 
6-73 


Aft 

3-3 


h 

2 








6 
3-56 


«-*>-> 






Polar Moment of Inertia J. 


Polar Modulus of Section — . 
c 


<—b- 


i 


bh 3 +hb 3 
12 


1 bh 3 -\-hb* 






6 Vb 2 +h 2 










6 








-. T 






A 

6 


1 

1 \ 


1 &i 4 -6 2 4 

3 v^ 






\ 




< 6^-> 




© 


ttJ 4 . 
32 


ird 3 

16 


( d - ) R 


o.ogS2(d 1 4 -d 2 i ) 


, rfl 4 -<*2 4 

0.1964 d 


Qi 


~(ba 3 +ab 3 ) 
64 


tv (ba 3 +ab 3 \ 
32 \ a / 



102 



MACHINE DESIGN. 
Table E. — Beams or Uniform Cross-Section 





Maximum 
moment 


Maximum 
deflection 


Cantilever, single load at end 


Wl 
\Wl 
\Wl 
\ Wl 

0.192 Wl 
\ Wl 
1 Wl 

1/12 Wl 


i WV 


Cantilever, uniform load 


3 EI 

1 Wl* 


Simple beam, load at middle 


8 EI 
1 Wl 3 




48 EI 
5 Wl 3 


Beam fixed at one end, supported at other, load near 
middle . . 


3HEI 

Q Wl 3 
. 0098 — 

Wl 3 

. 0054 -=T7 


Beam fixed at one end, supported at other, uniform 
load 


Beam fixed at both ends, load at middle 


J EI 

_i Wl 3 


Beam fixed at both ends, uniform load 


192 EI 

j_Wl 3 




384 EI 



W= total load, 1 = length, E = Young's modulus, / = moment of inertia. 
See further any handbook, or treatise on strength of materials. 



Table F. — Stress and Strain Formulae 



Unit stress in tension or compression 


p 

ft or fc= j 




Strain in tension or compression 


X=^ 




AE 


Unit stress in shear 


*P 

fs ~A 




Unit torsional stress 


. M t c 
fs=-j-, Mi=Pa 




Unit torsional stress, circular shaft 


, i6Mt 


Torsional strain 


QO i8oM*/ 
* = JG 




Unit stress in flexure 


, MbC 




Deflection in flexure 


See Table E 




Unit stress, combined flexure and ten. or comp. 


, P .M b c , . 
/=^jH — y~ (straight axis) 



PROPOR TIONS OF MACHINE PAR TS AS DICTA TED BY S TRESS. 103 

Table F. — Stress and Strain Formulae — Continued. 







Combined torsion and flexure 


Meb= 0.35^+0.65 VM b 2 +Mt 2 






Ditto, unit resultant normal tensile stress; cir. 
shaft 


ff _2(Mb+VMt*+Mb 2 




irr° 


- 




Ditto, unit resultant diagonal shear; cir. shaft, . 


,, 2VMt 2 +M b 2 
fs ~ vr* 






Combined tension and torsion 


/=o. 3 5/*+o.6 5 V/? +4/s 2 




Combined compression and torsion, long column 


7T 2 P Mt 2 
I 2 EI^~ 4 E 2 I 2 


Long Columns 


k= factor of safety 






One end free, other fixed 


P ir 2 EI min 




k 4 2 , 


Both ends free, guided in direction of load 


P ir 2 EI min 

k I 2 


One end fixed, other free but guided in direction 
of load 


P 2t 2 EI min 

k I 2 




Both ends fixed in direction 


P 4ir 2 EI min 




k I 2 



P = force, A = area, Z = length, M& = bending moment, Mt = torsional moment, — = 
J c 

section modulus, flexure,— = section modulus, torsion, E = modulus of elasticity, G- 
modulus of shear. c 



Table G— Stresses in Pressure Vessel Walls 

(ft = unit excess internal pressure, lbs. per sq. in.); (1 = thickness of plate, ins.). 



Thin cylinder, stress in long, section 



/*= 



D>p 

2t 



{D2— inner diam., in.) 



Thin cylinder, stress in cir. section. 



/*= 



D^p 

At 



Thin sphere, stress in any section , 



Ihp 

At 



Thick cylinder, stress in long. sec. Free 
ends. Di= external diam. (Max. strain 
theory) 



D 



\ iof— 1 



7p Birnie, and 
/— 13^ Grashof. 



Ditto, fixed or solid ends. (Max. strain 
theory.) 



A 



^ iof— 1 



Claverino. 



Ditto 


0.42^2 






Sames. 


Ditto, free ends. (Max. stress theory.) ... 


f _D,p 


Barlow. 


Ditto, either free or fixed ends. (Guest's 
max. shear law.) 


/ =2 ^( I_ 5;) 


Moss. 



104 



MACHINE DESIGN. 



Table G. — Stresses in Pressure Vessel Walls. — Continuel. 
Flat Plates, Uniformly loaded. 



Uniformly stayed, rows a inches apart 
fb= unit flexural stress 






Integral cast-iron cyl. head. R 2 = inner rad. 
r= rad. of inner curv. of flange 



/»^°- 8 r+\ 1 — 



p 

Bach. 



/>> 



Flanged steel, riveted cyl. head. 2? 2 and r 
as above. 0= \ to | 



o.5 r +* 



°-4+i) ) 



Flat circular unstayed plate, not flanged. 
0= 5 to f for cast-iron heads 
= f to | for free-lying steel heads 
= \ to | for st. heads, rigidly fastened at 

circumference 
= f to \ for st. heads fastened at cir. but 
yielding to equalize stress at center 



t>R 



H 



Concentrated Load at Center= P. 
(Plates supported, but not fixed, at edges). 



P 
Bach. 



Bach. 



and cir. 












Elliptical plate; a, major, b, minor axis. 
fh — | to -f cast iron 


■=»s/*4 r i 






Bach. 


Rectangular plate; a, major, b, minor side. 
<f> — f to It, cast iron 


Same as foregoing. 








Square plate; side= a. <j>= | to f, cast iron 


<=fV? 


Bach. 



Circular plate, rad. of \oa.d=R . 0=f, cast 
iron 


,>3./, 2 Ro\P 




Bach. 






Rectangular plate, a, major, b, minor side. . 
(f)=-^ to 2, cast iron 


b a 








Bach. 






Elliptical plate, a, major, b, minor axis. 
0= H to f , cast iron 




+3 
+3 


(-Y 

\aj b P 
Bach. 





CHAPTER VII. 



RIVETED JOINTS 



62. Methods of Riveting. — A rivet is a fastening used to unite 
metal plates or rolled structural forms, as in boilers, tanks, built- 
up machine frames, etc. It consists of a head, A, Fig. 56, and a 
straight shank, B. It is inserted, usually red-hot, into holes, 
either drilled or punched in the parts to be connected, and the 
projecting end of the shank is then formed into a head (see dotted 
lines) either by hand- or machine-riveting. A rivet is a permanent 
fastening and can only be removed by cutting off the head. A 
row of rivets joining two members is called a riveted joint or 
seam of rivets. In hand-riveting the projecting end of the 
shank is struck a quick succession of blows with hand hammers 
and formed into a head by the workman. A helper holds a sledge 





"^ 


„ 




*« 


mww 


WW 




A 


/ 



Fig. 56. 





Fig. 58. 



or ''dolly bar" against the head of the rivet. In "button-set" 
or "snap" riveting, the rivet is struck a few heavy blows with a 
sledge to "upset" it. Then a die or "button set," Fig. 57, is 
held with the spherical depression, B, upon the rivet; the head, 
A, is struck with the sledge, and the rivet head is thus formed. 
In machine-riveting a die similar to B is held firmly in the ma- 
chine and a similar die opposite to it is attached to the piston of 

105 



io6 MACHINE DESIGN. 

a steam, hydraulic, or pneumatic cylinder. A rivet, properly 
placed in holes in the members to be connected, is put between 
the dies and pressure is applied to the piston. The movable die 
is forced forward and a head formed on the rivet. 

The relative merits of machine- and hand-riveting have been 
much discussed. Either method carefully carried out will pro- 
duce a good serviceable joint. If in hand-riveting the first few 
blows be light the rivet will not be properly upset, the shank will 
be loose in the hole, and a leaky rivet results. If in machine- 
riveting the axis of the rivet does not coincide with the axis of 
the dies, an off-set head results. (See Fig. 58.) In large shops 
where work must be turned out economically in large quantities, 
machines must be used. But there are always places inacces- 
sible to machines, where the rivets must be driven by hand.* 

63. Perforation of Plates. — Holes for the reception of rivets 
are usually punched, although for thick plates and very careful 
work they are sometimes drilled. If a row of holes be punched 
in a plate, and a similar row as to size and spacing be drilled in 
the same plate, testing to rupture will show that the punched plate 
is weaker than the drilled one. If the punched plate had been 
annealed it would have been nearly restored to the strength of 
the drilled one. If the holes had been punched ~ inch to J inch 
small in diameter and reamed to size, the plate would have been 
as strong as the drilled one. These facts, which have been ex- 
perimentally determined, point to the following conclusions: 
First, punching injures the material and produces weakness. 
Second, the injury is due to stresses caused by the severe action 
of the punch, since annealing, which furnishes opportunity for 
equalizing of stress, restores the strength. Third, the injury 
is only in the immediate vicinity of the punched hole, since ream- 
ing out ^r inch or less on a side removes all the injured material. 

* See Sec. 75 for discussion of the importance of holding rivet under pressure 
until it is cooled; and the advantage of large rivets over small. 



For Single. Double, Treble Riveted Lap 'Joint 
For Single Riveted Double Butt Strap Joint 
. " Double ; " " " " " 

, " Treble " « " " " 




Thickness of Plates-T. 
Fig. 1. RIVET DIAMETERS BY BACH'S FORMULA. 





Fig. 3. SINGLE RIVETED LAP JOINT. 



Fig. 4- DOUBLE RIVETED LAP JOINT. 




^-^M 







Fig. 7. DOUBLE RIVETED DOUBLE BUTT STRAP JOINT. ' Fig.8, DOUBLE RIVETED 



PLATE I. 




1" Z'' '6" 4/' 5" 

Thickness of Plates-T. 

Fig. 2. RIVET LENGTH FOR GIVEN PLATE THICKNESS 





DOUBLE RIVETED LAP JOINT. 

1 



Fig. 6. TREBLE RIVETED LAP JOINT. 




RAP JOINT- Fig. 9. DOUBLE RIVETED DOUBLE BUTT STRAP JOINT* 



RIVETED JOINTS. 



107 



In ordinary boiler work the plates are simply punched and riveted. 
If better work is required the plates must be drilled, or punched 
small and reamed, or punched and annealed. Drilling is slow 
and therefore expensive; annealing is apt to change the plates 
and requires large expensive furnaces. Punching small and ream- 
ing is probably the best method. In this connection, Prof. A. 
B. W. Kennedy (see Proc. Inst. M. E., 1888, pp. 546-547) has 
called attention to the phenomenon of greater unit tensile strength 
of the plate along the perforations than of the original unperf orated 
plate.* Stoney (" Strength and Proportion of Riveted Joints," 
London, 1885) has compiled the following table: 



Table I. — Relative Percentage of Strength of Steel Plates Perforated 
in Different Ways. 



Specimens. 


Unit Strength of Net Section between Holes compared 
with that of the Solid Plate (ioo Per Cent). 




i Inch. 


i Inch. 


1 Inch. 


1 Inch, 


Punched 

Punched and annealed 

Drilled 


Per Cent. 
IOI .0 
105.6 
113. 8 


Per Cent 

94.2 

105.6 

Hi . 1 


Per Cent. 

82.5 

IOI .0 

106.4 


Per Cent. 

75-8 
100.3 
106. 1 



For punched and reamed holes the same percentages may 
be used as for drilled. 

Professor Kennedy gives constants which may be obtained 
from the following formula: Excess of unit strength of drilled 
steel plates in net section over unperforated section 
/ 6.i24\ /4.S -r\ 

H 2+ ^r)Hr) percent - 

/ is the thickness of plate in inches and r the ratio of pitch 
divided by diameter of hole. No data exist relative to iron 
plates in this matter. If r = 4.5, or more, there is no excess. 

64. Kinds of Joints. — Riveted joints are of two general kinds: 
First, lap-joints, in which the sheets to be joined are lapped on 

* This reported phenomenon is corroborated by tests made at Watertown 
Arsenal. See Tests of Metals, 1886, pp. 1264, 1557. It is fully explained by the 
condition of localized stress and the consequent prevention of lateral contraction. 



io8 



MACHINE DESIGN. 



each other and joined by a seam of rivets, as in Fig. 59a. Second, 
butt-joints, in which the edges of the sheets abut against each 
other, and a strip called a " cover-plate " or " butt-strap" is riv- 
eted to the edge of each sheet, as in c. In recent years a lap-joint 




]>©©©©© 

©©©©_©; 

© ©""©"""©""©"""©1 
©©©©©© 



k— Pitch— *T 



Fig. 60. 

with a single cover-plate has been used somewhat. It is shown 
in Fig. 592. 

There are two chief kinds of riveting: Single, in which there 
is but one row of rivets, as in Fig. 59a; and double, where there 
are two rows. 



RIVETED JOINTS. 



109 



Double riveting is subdivided into " chain-riveting," Fig. 596, 
and " zigzag " or " staggered " riveting, Fig. 59^. 

Lap-joints may be single, double chain, or double staggered 
riveted. 

Butt-joints may have a single strap as in c, or double strap; 
i.e., an exactly similar one is placed on the other side of the joint. 
Butt-joints with either single or double strap may be single, 



© 
© 




©©©©©©©© 

. J2L-JL -©=_-©_ JL -?L ®J?1 
r "©"""©""©"""©"©""©"^©"©] 

©©©©©©©© 




Fig. 60A 



double chain, or double staggered riveted. In butt-joints, single- 
cover-plates should have a thickness = / + \ 



and double cover- 



plates =-+- "; t being the thickness of the main plates. 



To sum up, there are: 

Lap-joints. . 



f Single-riveted 
Double chain-riveted 
Double staggered-riveted 



Butt-joints. 



f Single-riveted 

Single-strap -J Double chain-riveted 

[ Double staggered-riveted 

(Single -riveted 
Double chain-riveted 
Double staggered-riveted 



The demands of modern practice have added triple, quad- 
ruple and quintuple joints to the foregoing. In high-pressure 



no 



MACHINE DESIGN. 



cylindrical boilers, for instance, common practice is to employ 
for the longitudinal seam the highly efficient joint shown in 
Fig. 60. Here we have a triple-riveted butt-joint with double 
cover- plates; on each side of the joint two rows of rivets are in 



© 



© 

© © 



© 
© 



© 



© 
© 



©: 
©' 



©©©©©©©©©©©©© 
©©©©©©©©©©©©© 



©©©©©©©©©©©©© 
©©©©©©©©©©©©© 



L_® © 



© 



© 
© 



© 



© 
© 



©, 



-Pitch- 



FlG. 60B. 

double shear and one row, the outer, is in single shear. 
Quadruple and quintuple joints are shown in Figs. 60 A and 60B. 
65. Failure of Joint. — A riveted joint may yield in any one of 
four ways: First, by the rivet shearing (Fig. 61 a) ; second, by the 
plate yielding to tension on the line AB (Fig. 61 b) ; third, by the 
rivet tearing out through the margin, as in c; fourth, the rivet 
and sheet bear upon each other at D and E in d, and are both 
in compression. If the unit stress upon these surfaces becomes 
too great, the rivet is weakened to resist shearing, or the plate 
to resist tension, and failure may occur. This pressure of the 






Fig. 61. 



RIVETED JOINTS. Ill 

rivet on the sheet is called "bearing pressure." It is obvious 
that the strongest or most efficient joint in any case will be one 
which is so proportioned that the tendency to fail will be equal 
in all of the ways. 

66. Strength of Materials Used. — As a preliminary to the 
designing of joints it is necessary to know the strength of the 
rivets to resist shear, of the plate to resist tension, and of the 
rivets and plates to resist bearing pressure. These values must 
not be taken from tables of the strength of the materials of which 
the plate and rivets are made, but must be derived from experi- 
ments upon actual riveted joints tested to rupture. The reason 
for this is that the conditions of stress are modified somewhat in 
the joint. For instance, in single-strap butt-joints, and in lap- 
joints, the line of stress being the center line of plates, and the 
plates joined being offset, flexure results and the plate is weaker 
to resist tension, the rivets in the mean time being subjected to 
tension as well as shear; if the joint yield to this stress in the 
slightest degree the " bearing pressure " is localized and becomes 
more destructive. The effect of friction between the surfaces 
of the plates under the pressure at which they are "gripped" 
by the rivets is another item of considerable importance. Ex- 
tensive and accurate" experiments have been made upon actual 
joints and the results have been published.* 

The table on page 112 has been compiled as representing fair 
average results, and the values there given may be used for ordi- 
nary joints. 

The Master Steam Boiler-Makers' Assn., as the result of 
tests conducted by its committee, recommends, for iron rivets, 
/, = 42000 and f' s = 40000 ; for steel rivets, f s = 46000, f' s =44000. 

It will be noted that the values of f t are not given for steel. 

* See Proc. Inst, of Mech. Eng., 1881, 1882, 1885, 1888; Tests of Metals, 
Watertown Arsenal, 1885, 1886, 1887, 1891, 1895, 1896. Stoney's "Strength and 
Proportions of Riveted Joints," London, 1885. 



112 



MACHINE DESIGN. 



Table II. — Values of ft, /«, and } c for Different Kinds of Joints. 



Kind of Joint. 



Lap-joint, single-riveted, punched 
holes. 

Lap-joint, single-riveted, drilled holes 

Lap-joint, double-riveted, punched 
holes 

Lap-joint, double-riveted, drilled holes 

Butt-joints, single cover : Use values 
given for lap-joints. 

Butt-joints, double cover, single-riv- 
eted, punched holes 

Butt-joints, double cover, single-riv- 
eted, drilled holes 

Butt-joints, double cover, double-riv- 
eted, punched holes 

Butt-joints, double cover, double-riv- 
eted, drilled holes 

* Original plate 

* Original bar. 



Iron. 



40000 
45000 



45000 
50000 



40000 

45000 

45000 

50000 
50000 



38000 
36000 

40000 
38000 



u 

42500 

41000 
38000 
36000 



45000 



67000 
67000 

67000 
67000 



u 

89000 
89000 



89000 



Steel. 



60000 



47500 
45000 

48000 
46000 



48000 
46000 
47500 
45000 
52000 



85000 
85000 

85000 
85000 



u 

I 00000 



The tensile strength of steel varies through a considerable range 
due largely to differences in chemical constitution ; it also follows 
a rough law of inverse proportion to the thickness of plates; 
i.e., thin plates will be almost sure to show higher tensile strength 
than thicker plates of the same composition. Furthermore, the 
method of perforation greatly affects the strength of the plates, 
as has been pointed out in § 63. Ordinary boiler plates have a 
unit tensile strength ranging from 55,000 lbs. to 62,000 lbs. per 
square inch. For ordinary calculations f t may be taken as 
55,000 lbs. for punched plates and 60,000 for drilled plates. 
The shearing strength of rivets also varies inversely as their size, 
but these differences are slight. 

The Boiler Code Committee of the A. S. M. E. recommend: 
for iron rivets, f s = 38,000 and /',= 35,000; for steel rivets, 



* If the original material varies from this, the values given above should be 
varied proportionately. 



RIVETED JOINTS. 113 

/,= 42,000 and f' s = 39,000. They also recommend as the max- 
imum values to be used : for iron rivets, f s = 38,000 and f s = 38,000 ; 
for steel rivets, f s = 44,000 and /' s = 44,000. They recommend 
f t = 55,000 for mild steel, and/, = 45,000 for wrought iron, where 
the actual tensile strength of the plates is not known. Similarly 
for compressive strength they recommend / c = 95,000 for mild 
steel. 

67. Strength, Proportions, and Efficiency of Joints. — No 
riveted joint can be as strong as the unperforated plate. The 
ratio of strength of joint to strength of unperforated plate is 
called the joint efficiency. 

As stated in § 65 the highest efficiency for a joint is obtained 
when the relations between thickness of plate, diameter of rivet, 
pitch, and margin are such that the tendency for the joint to 
fail in any one way does not exceed the tendency for it to fail 
in any other way. Formulae can be developed for finding their 
proper values for each form of joint. 

Let d = diameter of rivet-hole in inches; t§-" > rivet; 
a = pitch of rivets in inches; 
/ = thickness of plates in inches ; 

ft = tensile strength of plates in pounds per square inch ; 
f e = crushing strength of rivets or plates, if rivets are in 

single shear, pounds per square inch ; 
//= crushing strength of rivets or plates, if rivets are in 

double shear, pounds per square inch; 
/,= shearing strength of rivets in single shear, pounds per 

square inch; 
//= shearing strength of rivets in double shear, pounds 

per square inch. 

Each joint may be treated as if made up of a successive series 
of similar strips, each unit strip having a width equal to a, the 
distance between centers of two consecutive rivets in the same 



ii4 



MACHINE DESIGN. 



row (see Fig. 62). If the stresses and . proportions for one such 
strip are determined, the results obtained will, of course, apply 
to all of the others, and consequently to the whole joint. Con- 
sider such a strip of thickness / and width a. 







5) o 
U o 



Fig. 62. 



Let P = ultimate tensile strength of unperforated strip, pounds; 
T= ultimate tensile strength of net section of strip, pounds; 
5= ultimate shearing resistance of all rivets in strip, 

pounds ; 
C = ultimate crushing resistance of all rivets or sides of 

holes, pounds; 
E= efficiency of joint. 
To illustrate this method, consider first the simplest joint, 
i.e., the single-riveted lap-joint. 

The unperforated strip has a tensile strength 



P=atj t 



(1) 



Along the row of rivets the net width of plate is less than the 
total width of the strip by an amount equal to the diameter of 
the rivet, and consequently the net tensile strength of the strip 
is expressed by the equation 

T = (a-d)tf t (2) 

In each unit strip there is but a single rivet with but one sur- 
face in shear, hence 



s =-/ 2 t° 



■7854^/. (3) 



RIVETED JOINTS. 1 15 

The crushing resistance of the rivet, or of the plate around 
the hole, may be written as 

C=dtj c (4) 

For highest efficiency T =S = C. 

Equating S and C, (3) and (4), . 78546^ =d// c . 

•'• d-i.2?J±. • • (5) 

This equation gives the proper theoretical value of d for a 
given value of t, and for materials represented by j c and / a . 
Equating T and S, (2) and (3), 

(a-d)tf t = . 7 Ss4d 2 }s- 

. .7854^/. ... 

•• a ~ — Jf t — + d ( 6 ) 

This gives the proper theoretical pitch. The efficiency of the 
joint is obtained by dividing T, S, or C by P. 

In most cases the values of d and a as determined by (5) 
and (6) cannot be strictly adhered to. Stock sizes of rivets 
must be used in practice, and there are also limitations connected 
with the largest sizes it is convenient to drive. These equations, 
furthermore, do not take into consideration the stresses set up in 
the rivets when their shrinkage, due to cooling, is resisted by 
the plates, an item which may become excessive with the smaller 
diameters. The spacing of the rivets must also be modified quite 
frequently by the proportions of the parts to be connected, by 
allowance for proper space to form the heads, and by provision 
for tightness. In practice it is therefore often necessary to depart 
from these values. 

It must be borne in mind, however, that any departure from 
the values of d and a given in (5) and (6) destroys the equality 



n6 MACHINE DESIGN. 

between T, 5, and C, and if such departure is made, the actual 
value of T, S, and C should be determined (by substitution of 
the values of d and a decided upon). The efficiency of the 
joint will then be found by dividing whichever has the smallest 
value, r, S, or C, by P. 

If the real efficiency of the joint is desired, the value of 
T must be obtained by increasing } t by the amount called for by 
the perforation of the plates. As explained in § 63 this will be 

/ 2 +~7= ) ( ~ ) per cent greater than the } t of the original, un- 

perf orated plate. 

68. Problem. — The following problem illustrates the method 
of using Table II in connection with the formulae (5) and (6). 

What should be the dimensions of rivet-hole and pitch for a 
single -riveted lap-joint for f-inch iron plates using iron rivets? 

Table II gives as values of f lt } 8 , and f c 40,000, 38,000, and 
67,000 lbs. per square inch, respectively, for this form of joint, 

< = r=-375"- 

Substituting these values, equation (5) becomes 



67000 . „ 

d = 1.27 X. 375 X~~ = .84" 

' °' D 38000 



and equation (6) 



. 7854 X. 84 X 38000 o . , 

a = J -^ — -7 + .84 = 2.24 inches. 

.375X40000 

69. Proportions of Single-riveted Lap-joints. — Table III 
and Table IV have been computed in this way. As Table IV 
refers to steel joints, the values of } t , j s , and } c are 55,000, 47,500, 
and 85,000 lbs. per square inch, respectively. 



RIVETED JOINTS. 



117 



Table III. — Proportions of Single-riveted Lap-joints, Iron Plates, and 
Rivets, Punched Holes. 





fc 


.7854^ 






t 


d=i.27t— 
IS 


<■= -is +d 


d 





3 
16 


.42 


1 . 12 






\ 




56 


1.49 






A 




70 


1.86 


f 


If 


1 




84 


2 . 24 


f-t 


lf-2 


h 


I 


12 


2.98 


f-xf 


2-2^ 


f 


I 


40 


3-73 


t-I 


2-2f 


3. 
4 


I 


68 


4-47. 


I-li 


2j-2f 


I 


I 


96 


5.22 


I-If 


2^-3 


I 


2 


24 


5-9 6 


I-I* 


2|-2f 


if 


2.52 


6.71 


lf-l| 


2f~3 



Column 1 gives the thickness of plate; columns 2 and 3 give 
the corresponding calculated values of d and a for joint of maxi- 
mum efficiency; columns 4 and 5 give the values of d and a 
as compiled by Twiddell in the Proc. Inst, of M. E., 1881, pp. 
293-295, from boiler-makers' practice. It will be noted that the 
rivets used in practice (see column 4) are considerably smaller 
in diameter than those called for in column 2, and that this differ- 
ence grows more and more marked as the thickness of the plate 
increases. The reason for this is that the difficulty in driving 
rivets increases very rapidly with their size, ij or if inches 
being the largest rivet that can be driven conveniently. The 
equality of strength to resist bearing pressure and shear is there- 
fore sacrificed to convenience in manipulation. As the diameter 
of the rivet is increased the area to resist bearing pressure in- 
creases less rapidly than the area to resist shear (the thickness 
of the plate remaining the same), the former varying as d and the 
latter as d 2 \ therefore if d is not increased as much as is neces- 
sary for equality of strength, the excess of strength will be to re- 
sist bearing pressure. If the other parts of the joint are made 
as strong as the rivet in shear, and this strength is calculated 
from the stress to be resisted, the joint will evidently be correctly 
proportioned. As machine-riveting comes into more general 



n8 



MACHINE DESIGN. 



use and pneumatic tools are used in "hand-work," this dis- 
crepancy will tend to disappear. 

Table IV. — Proportions of Single-riveted Lap-joints, Steel Plates, and 
Rivets, Punched Holes. 





tc 


.7854^5 






t 


's 


a =~ Tf— + d 


d 


a 


A 


•43 


1.08 


•47 


I* 


1 


•57 


i-45 


.61 


iA 


A 


•7i 


1. 81 


.81 


2 


I 


.86 


2. 17 


•94 


2& 


* 


1. 14 


2.89 


1. 19 


3 



Column 1 gives the thickness of the plate; columns 2 and 3 
give the values of d and a calculated for joint of maximum effi- 
ciency; columns 4 and 5 give proportions from practice, the 
authority being Moberly (see Stoney, " Strength and Proportions 
of Riveted Joints," p. 80). It will be noted how closely the 
theory and practice agree here for boiler joints .* 

70. Single-riveted Butt-joints. — To develop the general 
formulae for the values of a and d for single-riveted butt-joints 
with double cover-plates the same general method used in § 67 
applies. 

In this case the rivets are in double shear. Therefore 



4 



while T = (a — d)tf t , (2), as before and 

C=dtf c '. 
Equating S and C, (7) and (8), 



(7) 



(8) 



* For further data, compiled from American practice, see sec. 73. 



RIVE TED JOIN TS. 1 1 Q 

4 z 

fc' 

and d = .64j- / t (9) 

Equating T and S, (2) and (7), 

(a-d)tt t = i. S 7d 2 f/; 

/.a-^'+* do) 

For Double-riveted Lap-joints the unit strip contains two rivets, 
each in single shear. The following equations cover the case : 

T = (a-d)tf t , 



« xd 2 . 

S-2—.-I.S7&. 
4 



C = 2dtf c , 

d = i.2jjt (11) 

1.57^/, 
a = —Tf + d (12) 



71. Double-riveted Butt-joint. — For double riveted butt- 
joints, double cover-plate, either chain or staggered riveting, there 
are two rivets in double shear for each unit strip. 

T = (a-d)tf t , 
4 



120 MACHINE DESIGN. 

C = 2dtf c ', 

U 
J = .6477/. (13) 

Is 

.- 3 ^+*. ...... (M) 

72. General Formulae. — The following general equations for 
riveted joints have been developed by Mr. W. N. Barnard:* 

The unit strip is of width equal to the pitch, the maximum 
pitch being taken unless all rows have the same pitch. 

The general expression for the net tensile strength of the unit 
strip is 

T=(a-d)tf t (15) 

The general expression for resistance to shearing of the rivets 
in the unit strip is 

„ nnd 2 , 2mnd 2 . , 

4 4 

in which n equals the number of rivets in single shear and m 
equals the number of rivets in double shear. 

The general expression for resistance to crushing of the unit 
strip is 

C = ndtf c +mdtf/ (17) 

The tensile resistance of the solid strip is 

P = atf t (18) 

Equating 5 and C, (16) and (17), and transposing, we get 

nf -\-mf r 

*-*■**£& • (I9) 

* See article, " General Formulas for Efficiency and Proportions for Riveted 
Joints," by Professor J. H. Barr in Sibley Journal of Engineering, Oct., 1900. 



'RIVE TED JOIN TS. 1 2 1 

Equating T and C, (15) and (17), 

a= ^L±Miy +d (2o) 



/ 

Or, equating T and S, (15) and (16), 



a = .7&5#P{ t < )+ d. . . . (21) 

The following equation for efficiency has been developed 
on the assumption that T = S = C. 

From E = — we get, by substitution and transposition, 

£ = - f • • (22) 

I+ h 

nfc+mfj 

This equation is useful in finding the limiting efficiency of 
joint for any form and materials; the actual proportions adopted 
may give a lower efficiency, but can never give a higher efficiency.* 
73. Proportions of Joints. — In American practice it will be 
found that there is more or less departure from the proportions 
which would be arrived at by the strict application of the prin- 
ciples laid down in the preceding articles. This variation is 
due to several considerations. Chief among them is the practical 
difficulty of driving large rivets, thus leading to the adoption of 
rivet diameters with reference to convenience of manipulation 
rather than efficiency of joint. As machines displace handwork 
the reason for this departure disappears and there is an increasing 
tendency to use the larger and more correct rivet diameters. 
Conservatism must be reckoned with here and also in the failure 
to recognize the fact that rivet diameters do not depend solely 

* In the Proceedings of the Inst, of M. E., 1881, there is an article entitled 
"On Riveting, with Special Reference to Ship-work," M. Le Baron Clauzel, 
which enters deeply into the development of general formulae. 



122 MACHINE DESIGN. 

upon the thickness of plates, but also should vary with the kind 
of joint. Practice tends to hold to one diameter of rivet for each 
thickness of plate, irrespective of the kind of joint. 

Another item of practical importance is tightness against 
leakage under pressure. Most formulae are developed without 
consideration of this important factor. From a practical point 
of view, the joint fails when it begins to leak; actual rupture 
need not take place. The topic of the allowable maximum 
pitch as governed by experience with tightness of joints is 
discussed in § 80. 

The margin in a riveted joint is the distance from the edge of 
the sheet to the rivet hole. This must be made of such value that 
there shall be safety against failure by the rivet tearing out. 
There can be no satisfactory theoretical determination of this 
value; until recently it has been held that practice and experi- 
ments with actual joints showed that a joint would not yield in 
this way if the margin were made = d = diameter of the rivet 
hole. This is a safe rule for iron rivets in steel plates for any 
type of joint. Where steel rivets are used it will be well to 

increase this to — d. 
4 
The American Machinist, May 3, 1906, says: The minimum 

distance from the center of any rivet hole to a sheared edge 
ought not to be less than i\" for J" rivets, if" for f " rivets, if" 
for f" rivets, 1" for -|" rivets; and to rolled edges if", if", 1", 
and |", respectively. The maximum distance from any edge 
should be eight times the thickness of plate. 

The distance between the center lines of rows may be taken 
not less than 2.5^ for double-chain riveting, and 1.88^ for 
double-staggered riveting. This will insure safety against zig- 
zag tearing of the plate, but brings the heads very close together. 
From these values and those of margins, as just discussed, the 
proper amount of lap can readily be determined for any kind 
of joint. 



RIVETED JOINTS. 



123 



74. Relative Efficiencies of Various Kinds of Joints. — The 

actual efficiencies of joints when tested show some departure 
from the calculated ideal efficiencies. The following Table (V) 
has been compiled from the results of tests to show roughly the 
relative efficiencies of various types of joints : 

Table V. — Relative Efficiency of Iron Joints. 



Efficiency 
Per Cent. 



Original solid plate 

Lap-joint, single-riveted, punched 

drilled 

double " 

Butt-joint, single cover, single-riveted. , 

" " double-riveted 

double ' ' single-riveted . 

" " double-riveted 



45 
5o 
60 

45-5o 
60 

55 
66 



Relative Efficiency of Steel Joints. 



Efficiency Per Cent. 



Thickness of Plates. 

i-i *-* l-t 



Original solid plate 

Lap-joint, single-riveted, punched 

drilled 

' ' double-riveted, punched 

drilled 

Butt-joint, double cover, single-riveted, drilled . . 

" double-riveted, punched. 

drilled.., 



100 

5o 
55 
75 
80 
70 

75 
80 



100 
45 
50 
70 

75 
65 
70 

75 



100 
40 
45 
65 
70 
60 

65 

70 



These tables are from Stoney's "Strength and Proportions of Riveted Joints." 



Triple riveted butt-joints with double cover-plates show 
efficiencies ranging from 80 to 90 per cent.* 

Quadruple joints, of this form, range from 90 to 95 per cent; 
and quintuple, from 95 to 98 per cent. 



* For details of joints tested, see Tests of Metals, Watertown Arsenal, 1896. 



124 MA CHINE DESIGN. 

75. Slippage. — At about 25 to 35 per cent of its ultimate load 
slippage takes place in a riveted joint. This is probably due 
to the fact that at this load the friction between the plates, owing 
to the pressure exerted on them by the rivets, is overcome. It 
has been found the larger the cross-sectional area of the rivet 
the greater the percentage of ultimate load which can be with- 
stood without slippage. It has also been found that large rivet- 
heads are better than small ones for the same reason. 

The importance of the consideration of slippage has been fully 
established by the work of Professor Bach (" Die Maschinen- 
elemente," 9th ed. pp. 164-195). His careful and exhaustive 
experiments prove that : 

1. In cooling the rivet shrinks away from the walls of the 
hole. 

2. In consequence of this, there is no tendency to shear off 
the rivet until after the joint has failed, for all practical purposes, 
by losing tightness because of slippage. 

3. The percentage of the ultimate or rupture load at which 
slippage takes place varies according to three items : 

a. It is directly proportional to the square of the diameter 

of the rivet. From this the desirability of using large 
rivets rather than small is further established. 

b. It is increased by calking, especially if both rivet heads 

are calked as well as the plate edges. 

c. It is greatly increased by holding the rivets under maxi- 

mum pressure until they are cool enough to have set. 

This gives better results than blows, light pressure, 

or early removal of pressure. 
Professor Bach argues that joints should not be proportioned 
with reference to the ultimate or rupture strength. He claims 
that the maximum pitch is determined by the condition of tight- 
ness against springing open between rivets when pressure is 
applied. The minimum pitch is that fixed by the spacing 
of rivet heads which is the least which will permit calking them. 
Between these limits he chooses pitch: 



RIVETED JOINTS. 



125 



i. So that the safe resistance to slippage (as experimentally 
determined by him) is equated to the stress on the joint due to 
the diameter of the vessel and the pressure. 

2. So that the unit stress in the plate between the rivets shall 
not exceed the safe working value of the plate material when the 
strength of the perforated section is equated to the stress due to 
the diameter and pressure. 

Plate I shows graphically the proportions of riveted joints as 
determined from Professor Bach's formulas by M. Shibata in 
the American Machinist, Vols. 26 and 27. 

76. Rivet Size and Proportions. — In general the rivet should 
have a shank A inch smaller in diameter than the hole to be 
filled, while the head should have a diameter of from 1.6 to 2 
times the diameter of hole, and a height of from .6 to .75 times 
the hole diameter. Especial care should be taken in the case 
of machine-riveting to have just enough metal projecting beyond 
the hole to allow for the necessary upset for the shank to fill 
the hole, with just enough left over to fill the die for the head. 



^~ 




t^-E^f 



d 



""' 



. — E -v 
A B C 

Fig. 63. 
TABLE OF DIMENSIONS OF RIVET HEADS. 



Diameter 




Pan Head. 




Button Head. 


Counter Sunk. 


of Rivet. 




A 




B 


c 


d 


E 


F 


G 


E 


G 


E 


G 


1 


lA 


1 9 
"3 2- 


A 


I* 


A 


lA 


A 


ft 


li 


to 


If 


I* 


* 


I* 


A 


I 


li 


fi 


H 


li 


& 


li 


I 


I 


If* 


*f 


f 


I* 


3. 

4 


If 


7 


1 


If 


if 


h 


If 


f 


If 


i 



126 MACHINE DESIGN. 

77. Problem. — How far must the tail of the rivet project in 
order to satisfy the above conditions for the following case: Two 
plates f inch thick each are to be connected, using J-inch rivets 
in H-inch holes. The head is to be cone-shaped, having an out- 
side diameter of if inches and a height of f inch. 

The cubical contents of the cone head = area of base X J 
altitude = 2.76 square inches X. 2 5 inch = .69 cubic inch. 

The difference in cubical contents between a hole xf inch in 
diameter by f inch long and a shank J inch in diameter and 
I inch long = |X.7854(i| 2 — J 2 ) = .o67 cubic inch. 

The amount required for head and upset therefore equals 
.69 + .067 = .757 cubic inch. 

The area of the f -inch shank = .60 square inch. .757 cubic inch, 

•757 
therefore, calls for a length of -7— = 1.25 inches. This amount 

would then be the projection through the plate. The length of 
rivet-shank called for would equal f inch + 1 J inches = 2 inches. 

Note. — Had the head been cup-shaped, its cubical contents 
should have been taken as that of a spherical segment. For cup- 
shaped heads the diameter is about 1.7 X diameter of hole, and 
the height about .6 X diameter of hole. The volume of the 
spherical segment is given by the following rule: Multiply half 
the height of the segment by the area of the base and the cube 
of the height by .5236 and add the two products. 

78. Countersunk Rivets. — Fig. 63 C shows the proportions for 
a countersunk rivet. C6untersunk rivets make a much weaker 
and less reliable joint than the ordinary form, and should only 
be used where it is absolutely necessary that the surface of the 
plate be free from projections. 

79. Nickel-steel Rivets. — Where peculiar conditions call for 
great strength of rivet combined with small area, it may be found 
desirable to use nickel-steel rivets. Experiments made by Mr. 
Maunsel White (see Journal Am. Soc. of Nav. Eng., 1898) on 



RIVETED JOINTS. 127 

riveted joints using nickel-steel rivets showed an average shear- 
ing resistance of 85,720 lbs. per square inch for single shear, and 
an average of 90,075 lbs. per square inch for double shear. These 
values, it will be noted, are nearly double those of the very mild 
steel ordinarily used. The rivets were § inch in diameter, and 
some of the joints failed by tearing the plates, while others failed 
by shearing the rivets * 

80. Construction of Tight Joints. — In general three types 
of riveted joints may be recognized : 

1. Those in which strength is the sole factor of importance, 
as in most purely structural iron and steel work. 

2. Those in which strength and tightness are equally deter- 
mining elements, as in boilers and pressure pipes. 

3. Those in which tightness is the prime consideration, as 
in tanks subjected to only light pressure. 

In punching holes in plates, it is, of course, necessary to 
have the hole in the die-block larger than the punch. The 
comequence is that the holes are considerably tapered and 
care should be exercised in joining the plates that the 
small ends of the holes be together as shown in Fig. 64A, 
and not apart as shown in Fig 64B. It is obvious that at 
A the pressure on the rivet tends to draw the plates closer 
tog.ther, and that as the rivet cools its lorn itudinal shrinkage 
will tend to keep it a tight fit for the hole in spite of its diametral 
shrinkage. 

It is equally obvious that at B the pressure on the rivet will 



* See Bulletin No. 49, Eng. Exp. Station, University of Illinois, for exhaustive 
tests, by Professors Talbot and Moore, on nickel-steel riveted joints, undertaken at 
the request of the Board of Engineers of the Quebec Bridge and the Pennsylvania 
Steel Co. The chief conclusions to be drawn are that, while these joints show 
advantage over ordinary carbon steel as concerns ultimate strength, there is no 
such advantage as regards slippage. All of the joints tested failed by shearing the 
rivets. This shows that the rivets were theoretically too small and accounts for 
the slippage. See Sec. 75. 



128 MACHINE DESIGN. 

tend to force the plates apart and squeeze metal between them, 
and also that all shrinkage of the rivet will be away from the 
walls of the hole. 

In using drilled plates care must be exercised to remove the 
sharp burrs left by the drill, as experience has shown that this 
has a considerable effect on the strength of the joint. 

Where the plates form the walls of vessels to hold fluids, the 
joints must be designed with a view toward tightness as well as 




ItH f b MSa t ? —? -* 

Fig. 64. Fig. 65. 

strength. For this purpose the edges are planed at a slight bevel, 
and calked as shown in Fig. 65 by a tool which resembles a cold- 
chisel with a round nose. Pneumatic tools are used for this pur- 
pose almost entirely, as they execute more uniform and rapid 
work than can be done by hand. In calking great care should 
be exercised not to groove the plates at A- A, as these are danger- 
points for bending, and an incipient groove is very apt to develop 
into a crack. It is largely on this account that the round- 
nose calking-tool has superseded the square-nose in the best 
practice. 

It has been found that the load which the joint will carry 
before leaking is greatly increased if both rivet heads are calked 
as well as the plate edges. (Bach's experiments.) 

The consideration of tightness has a determining effect on the 
maximum allowable pitch for any given thickness of plate and 
type of joint. Based upon practice the following values have 
been found safe for r 5 ^ " plates: — 

Single riveted lap joints, pitch = 7t 

Double riveted lap joints, pitch = 9-5t 

Double riveted butt joints, pitch (in outer row) = I4«5t 

Triple riveted butt joints, pitch (in outer row) = 2ot. 



RIVETED JOINTS. I2Q 

Because of the use, with heavier plates, of rivet diameters 
which are proportionately too small, these ratios of pitch to 
thickness of plate will be found to decrease in practice as the 
thickness of plate increases. Thus for J" plates they become 
5t, 6.6t, io.25t, and i6t, respectively.* 

81. Materials to be Used. — The material to be used in 
riveted joints depends, of course, on the nature of the work, 
but in general it may be said that extremely mild and highly 
ductile steel as free from phosphorus and sulphur as possible 
should be used. Open-hearth steel is greatly to be preferred to 
Bessemer, f 

82. Plates with Upset Edges. — Some boiler-makers have 
adopted, as an expedient for saving material, a method of using 
plates with thickened (upset) edges. If we let t represent the 
thickness of the body of the plate and f the thickness of the edge, 
while a represents the pitch and d the diameter of hole, then, when 

a 

the joint will be as strong as any other section of the plate, the 
joint being proportioned, of course, for the thickness f. It is 
customary to thicken only the edges which form the longitudinal 
seam. This method is open to two serious objections. Unless 
the plates are very carefully annealed after being upset they are 
almost certain to be weakened by indeterminate working and 
cooling stresses. Moreover, although the original new joint may 
show as high an efficiency as if the plates throughout were of the 
thickness f, as corrosion proceeds, it acts more on the plate away 
from the joint than at the joint, because at the latter place the 
plate is protected by the cover-plate or rivet-head or both. The 

* For quadruple and quintuple butt joints they may be 32/ and 48/, respect- 
ively. 

t Standard specifications can be found in the A.S.M.E. Boiler Code. 



13° 



MACHINE DESIGN. 



result is a shorter life under full pressure for the boiler with thin 
plates and thickened edges. 

83. Joints for More than Two Plates. — The joints considered 
thus far have dealt with the problem of connecting the edges 
of two plates only. In tanks and boilers which must have tight 
seams we are frequently confronted with the problem of joining 
three and even four plates. An instance is where the cross-seam 
and longitudinal seam of a boiler meet. The joint is made by 
thinning down one or more of the plates. Figs. 66 to 70 (taken 
from Unwin's "Machine Design") show the methods employed. 
Fig. 66 shows a junction of three plates, a, b, and c, where both 
seams are single- riveted lap-joints. It will be seen that the 
corner of a is simply drawn down to an edge and '* tucked 
under" c. 

Fig. 67 shows a junction of three plates where one seam is a 
single-riveted and the other a double-riveted Jap-joint. As 
before, the corner of a is drawn down and tucked under c. 

Fig. 68 shows the junction of three plates where both seams 
are single -riveted, single cover- plate butt-joints. The plates 
merely abut against each other, but the longitudinal cover is 





Fig. 66. 



Fig. 68. 



Fig. 67. 

drawn down and tucked under the cross-seam cover which is 
thinned down to match. 

Fig. 69 also shows the junction of three plates ; here the cross- 
seam is a single-riveted lap-joint while the longitudinal joint 
is a double-riveted butt-joint with double cover-plates. The 



RIVETED JOINTS. 



131 



upper cover-plate is planed on the end so that it can be tightly 
calked where it abuts against the plate c. 





Fig. 69. 



Fig. 70. 



A method of joining four plates is shown in Fig. 70. Both 
seams are single-riveted lap-joints, b and c are both drawn 
down as shown. 

84. Junction of Plates Not in Same Plane. — Where the plates 
to be joined are in different planes, it is customary to use some 
one of the rolled structural forms. Fig. 71 shows the method 
of using an angle iron for plates at a right angle to each other. 

Where it is possible to turn a flange on one of the plates this 
method is often adopted. Care should be taken not to use too 



0000 

— £=* ^-n •-n 



Fig. 71 





Fig. 72. 



sharp a radius of curvature (the inside radius must be greater 
than the thickness of the plate even with the mildest steel) and 
the flanged plate should be thoroughly annealed after it is bent. 

Fig. 72 shows the method of making flanged joints such as 
are frequently used in connecting boiler heads and shells. 

85. Problem. — The following problem will serve to illustrate 
the design of riveted joints for boilers. It is required to design a 



132 MACHINE DESIGN. 

horizontal tubular boiler 48 inches in diameter to carry a work- 
ing pressure of 100 pounds per square inch. 

A boiler of this type consists of a cylindrical shell of wrought 
iron or steel plates made up in length of two or more courses or 
sections. Each course is made by rolling a flat sheet into a 
hollow cylinder and joining its edges by means of a riveted joint, 
called the longitudinal joint or seam. The courses are joined to 
each other also by riveted joints, called circular joints or cross- 
seams. Circular heads of the same material have a flange turned 
all around their circumference, by means of which they are 
riveted to the shell. The proper thickness of plate may be 
determined from (I) The diameter of shell = 48 inches ; (II) The 
working steam-pressure per square inch = 100 pounds; (III) The 
tensile strength of the material used; let steel plates be used 
of 60,000 pounds specified tensile strength. 

Preliminary investigations of the conditions of stress in the 
cross-section of material cut by a plane (I) Through the axis; 
(II) At right angles to the axis, of a thin hollow cylinder, the 
stress being due to the excess of internal pressure per square inch. 
Let / =the length of the cylindrical shell in inches; 

Z) = the diameter of the cylindrical shell in inches; 

^ = the excess of internal over external pressure in pounds 
per square inch; 

/i=unit tensile stress in a longitudinal section of material 
of the shell due to p; 

J2 = unit tensile stress in a circular section of material of the 
shell due to p ; 

/ = thickness of plate; 

ft = ultimate tensile strength of plate. 
All stresses are in pounds per square inch. 

In a longitudinal section the total stress is equal to lDp t 

Dp 

and the area of metal sustaining it = 2IL Then f\= — - . 

2,t 



RIVETED JOINTS. 133 

nD 2 p 

In a circular section the total stress = — — , and the area 

4 

sustaining it =nDt } nearly. Then 

1 - n 2 2 J.^ £P 

h ~ 4 X *Dt~ 4 f 

Therefore the stress in the first case is twice as great as in 

the second; and a thin hollow cylinder is twice as strong to 

resist rupture on a circular section as on a longitudinal one. 

The latter only, therefore, need be considered in determining the 

thickness of plate. Equating the stress due to p in a longitudinal 

section, and the strength of the cross-section of plate that sustains 

Dp 
it, we have lDp = 2ltj t . Therefore t=~r, the thickness of plate 

that w T ould just yield to the unit pressure p. To get safe thickness, 
a factor of safety K must be used. It is usually equal in boiler- 
shells to 5 or 6. Its value is small because the material is highly 
resilient and the changes of pressure are gradual, i.e., there are 
no shocks. This takes no account of the riveted joint, which 
is the weakest longitudinal section, E times as strong as the 
solid plate, E being the joint efficiency =0.75 if the joint be 

KDp 
double-riveted. The formula then becomes t= , „ . Sub- 

2f t E 

stituting values, 

6X48X100 . 1 . _ 

/ = -: — =0.32 inch, say & inch. 

2X60000X0.75 ° ' J 

The circular joints will be single-riveted and joint efficiency 
will =0.50. But the stress is only one half as great as in the 
longitudinal joint, and therefore it is stronger in the proportion 
0.50X2 to 0.75, or 1 to 0.75. From this it is seen that a circular 
joint whose efficiency is 0.50 is as strong as the solid plate in a 
longitudinal section. From the value of / the joints may now be 
designed. 



134 MACHINE DESIGN. 

Consider first the cross-seam. This is a single-riveted lap- 
joint. Assume drilled holes. 
Equations (5), 

Is 

.7854^/3 , 
and (6), a= / ^ ' + d, 

apply, while from Table II we get as values of } t , / 8 , f c , 60,000, 

45,000, and 85,000 pounds per square inch respectively for steel 

plates and rivets. 

, 8qooo . 

.*. d = i.2jX— X.3i25=.75inch 

' 45000 ° J ,D 

— 2 
•78q4X.7^ X45000 „ . , 

and a = — 7-7 + .75 = 1.81 inches, say iH inches. 

.3125X00000 lo ' J : 

The margin = d = .75 inch. 

The lap of the cross-seam =$d = 2.25 inches. 

The longitudinal seam will be a double staggered lap-joint. 

Equations (11) and (12) apply: 

d -1.27k and a=^-t s + d. 

Is Hi 

From Table II, } t =60,000, } a =46,000, and } c =85,000; 

, 85000 . . 

'"" J==I * 2 7 x ^6o^o X,3I25==,74mC ,Say * 75, 



2 

i. 57 X. 75 X46000 
and a = <3I25x6oooo + -75 = 2.93 inches, say 2 ft inches. 

The distance between the rows = 1.88^ = 1.41 inches, say i& 
inches. The total lap in the longitudinal joint =4.88^ = 3.66 
inches, say 3^ inches. 



RIVETED JOINTS. 135 

The joints are therefore completely determined, and a detail 
of each, giving dimensions, may be drawn for the use of the work- 
men who make the templets and lay out the sheets. 

Having determined the proportion of the joints, let these 
dimensions be used to calculate the actual efficiency of the longi- 
tudinal seam. 

Assume that the natural tensile strength of the unperforated 
plate is 60,000 lbs. per square inch. 

The excess of strength of drilled steel plates in net section 
over unperforated section (see § 63) is 



6.124A /4-5— r N 

, 2+ vTA~' per 



a 2 -9375 
Here / = .3i25m. and r = ~T == ~ 1 — =3-9 2 ; 

.\ excess due to perforation =3%, nearly. 
60,000X1.03 =61,800 = /*. 
/, =46,000 from Table II. 

c =85,000 

T = (a-d)tf t = (2.9375 - .75)03125 X6i,8oo) =42,250 lbs. 
S = i.57d 2 / s = i. 57 X-75 2 X46,ooo =40,625 lbs. 
C = 2dtj c = 2 X. 75 X. 3125X85,000 = 39,845 lbs. 
P = a/ ( / = 2.9375X6o,oooX.3i25 =55,075 lbs. 
Of T, S, and C, the latter has the smallest value; the actual 

efficiency of the joint may be taken as = p = = -7 2 35 or 

72-35%- 

Since T, S, and C are unequal it is evident that there has been 
departure from the conditions for maximum efficiency. There are 
two ways of restoring this equality, or at least diminishing the 
inequality. If a be slightly decreased, Tand P will be propor- 



I3 6 MACHINE DESIGN. 

tionately decreased and S and C will have the same values as 
before. 

Leaving a as before and increasing d, increases S as the 
square of d and C as d, while T and P remain as before. 

Inspection shows that T exceeds C by 5.7 per cent. There- 
fore a may be decreased by this percentage, or .17 inch. This 
is approximately & inch and reduces the pitch from 21! to 
2f inches. 

Using this value of a gives as the excess strength due to per- 
foration 4.33 per cent. 

.*. } t = 62,600 lbs., 

T = 39,i25 lbs., 
5=40,625 lbs., 

C = 39> 8 45 lbs., 
P = 51,560 lbs., 

P 51560 /:) /u 

The second method of balancing T, 5, and C would be by 
increasing d. The next commercial size above f inch would 
be H inch. Leave a = 2ft inches, and increase d to H inch, and 
first calculate the excess strength due to perforation. 

/ = -3i25 inch, r=-j-= * . 
00 ' rf .8125 

Using these values, the excess = 4. 5 7 per cent. 



and, since T is least, 
E 



n- 


= 62,750 


lbs., 


T- 


= 41,670 


lbs., 


S = 


= 47^75 


lbs., 


c- 


= 43> l6 5 


lbs., 


p= 


= 55^75 


lbs., 


T 
P = 


41670 

".S.S075" 


= 75.66%, 



RIVETED JOINTS. 137 

The best result is that obtained by keeping d = \ inch, but 
changing a to 2J inches, and it would be advantageous to make 
these the proportions of the joints rather than those first de- 
termined. 

The next step is to check back, using the proportions de- 
cided upon, for the actual factor of safety which should not be 
less than 5. 

KDp 
From the equation (p. 133) t=—r^ , we have 

K 2ftEt 

m 2X62600X.7588X.3125 
.. A = 5— = 0.IO. 

48 X 100 

Attention should be called to the fact that the ordinary and 

less correct method of calculating efficiencies ignores the excess 

strength due to perforation, and the efficiency is simply taken 

T 
as y. 

With a = 2$ inches, and d = f inch, this would give us 

£ = ^75 = 74.47%> instead of 72.35%. 
With a = 2f inches and d = \ ip<:h, it would give us 



375oo 
51560 



E= \ =72.71%. instead of 75.oo 70 . 



With a = 21$ inches and d=H inch, it would give 
£ = 55^75 = 7I ' 97% inSt6ad ° f 75 ' 66%# 



138 MACHINE DESIGN. 

In practice the designer must be familiar with the code of 
rules, governing all details of boiler design, which prevails where 
the boiler is to be used. There are many such codes — private, 
state, and national, and they are not in agreement. 

One, recently prepared by a committee of the American Society 
of Mechanical Engineers, will, it is to be hoped, supersede 
them all for the entire United States. 



CHAPTER VIII. 



BOLTS AND SCREWS. 



86. Classification and Definition. — Bolts and screws may be 
classified as follows: I. Bolts; II. Studs; III. Cap-screws, or 
Tap-bolts; IV. Set-screws; V. Machine screws; VI. Screws 
for power transmission. 

A "bolt" consists of a head and round body on which a 
thread is cut, and upon which a nut is screwed. When a bolt 
is used to connect machine parts, a hole the size of the body of 
the bolt is drilled entirely through both parts, the bolt is put 
through, and the nut screwed down upon the washer. (See 
Fig- 73-) 




Fig. 73. 



Fig. 74. 



Fig. 75. 



A "stud" is a piece of round metal with a thread cut upon 
each end. One end is screwed into a tapped hole in some part 
of a machine, and the piece to be held against it, having a hole 
the size of the body of the stud, is put on and a nut is screwed 
upon the other end of the stud against the piece to be held. (See 
Fig. 74.) 

139 



140 MACHINE DESIGN. 

A " cap-screw" is a substitute for a stud, and consists of a 
head and body on which a thread is cut. (See Fig. 75.) The 
screw is passed through the removable part and screwed into a 
tapped hole in the part to which it is attached. A cap-screw 
is a stud with a head substituted for the nut. 

A hole should never be tapped into a cast-iron machine 
part when it can be avoided. Cast iron is not good material 
for the thread of a nut, since it is weak and brittle and tends to 
crumble. In very many cases, however, it is absolutely neces- 
sary to tap into cast iron. It is then better to use studs if the 
attached part needs to be removed often, because studs are put 
in once for all, and the cast-iron thread would be worn out 
eventually if cap-screws were used. 

The form of the United States standard screw-thread is 
shown in Fig. 76. The sides of the thread make an angle of 

6o°. Instead of coming to a sharp 
_ r 7\ 6<k/\ point, the threads have a flat at top 
and bottom whose width is = J^, p 
'V % being the pitch. Table VI gives the 

Fig. 76. standard proportions. 

For single threads the lead of the thread helix equals ^>,for double 
and triple threads it equals 2p and 3^, respectively. If clockwise 
rotation of the screw causes the thread to enter the nut, the thread 
is termed right-hand; if counter-clockwise, left-hand. 

When one machine part surrounds another, as a pulley -hub 
surrounds a shaft, relative motion of the two is often prevented 
by means of a " set-screw," which is a threaded body, pref- 
erably non-projecting (Fig. 77). The end is either rounded as 
in Fig. 77 a, or pointed as in Fig. 77 b, or cupped as in Fig. 
77 c, and is forced against the inner part by screwing through 
a tapped hole in the outer part. 

Data relative to the holding power of set-screws will be 
found in § 109. 




BOLTS AND SCREWS. 

Table VI. — U. S. Standard Screw-threads. 



141 







Bolts and Threads. 




Hex. Nuts and Head: 




Sq. 
iVandi*. 


*j 


jd 









"o 


u 


V 


CD 


•a 






"o 


















S3 






n 


i_ 


tf 
















6 




O 






"o 


<u 


*°^ 


£ 




Pi 


rt 


cfl 


cfl 






u 


ex 

e/J 


Ss 


O 


"o . 


^^ 


Bj 


S • 

r! 


a* 




0) 


a*- 


g 


T3 
a) 


£h 


A 


rt 


g 


bo 

1- 




bo 
bp§ 


G 


G 


bog 


a 


JC 


.S'S 


•d 
k 


£W 


%£ 


.§* 


.§£ 


g* 


15 


!5 


ga 


5 


P 


p 


< 


< 


CO 


CO 


J 


H 


H 


►J 


Ins. 




Ins. 


Ins. 


Sq. Ins. 


Sq. Ins. 


Ins. 


Ins. 


Ins. 


Ins. 


Ins. 


Ins. 


i 


20 


•185 


.0062 


.049 


.027 


i 


A 


H 


1 


A 


7 
T7F 


A 


18 


.240 


.0074 


.077 


•045 


ft 


ii 


ft 


A 


i 


T§ 


1 


16 


.294 


.0078 


. no 


.068 


ii 


f 


¥ 


1 


A 


63 


A 


14 


•344 


.0089 


.150 


•093 


If 


§1 


Ttf 


A 


f 


i,V 


i 


13 


.400 


.0096 


. 196 


. 126 


1 


if 


I 


i 


A 


lif 


A 


12 


•454 


.0104 


.249 


. 162 


fi 


If 


Ii 


A 


i 


ill 


1 


II 


•5°7 


.0113 


• 307 


.202 


iA 


1 


iA 


f 


A 


Ii 


i 


IO 


.620 


.0125 


.442 


.302 


ii 


iA 


iA 


f 


ft 


ill 


i 


9 


•73i 


.0138 


.601 


.420 


iA 


if 


ift 


i 


ft 


2A 


1 


8 


•837 


.0156 


• 785 


•550 


if 


iA 


ii 


1 


ft 


a« 


il 


7 


.940 


.0178 


•994 


•694 


ill 


if 


2& 


ii 


iA 


2A 


ii 


7 


1.065 


.0178 


1.227 


•893 


2 


ift 


2A 


ii 


iA 


aff 


if 


6 


1. 160 


.0208 


1.485 


1-057 


*A 


2i 


2ft 


if 


iA 


3A 


ii 


6 


1.284 


.0208 


1.767 


I.295 


2 t 


2A 


2f 


ii 


iA 


3tf 


if 


si 


1.389 


.0227 


2.074 


I-5I5 


2A 


2* 


2§i 


if 


iA 


3l 


if 


5 


1. 491 


.0250 


2.405 


I.746 


2f 


4i 


3A 


if 


ift 


3ff 


if 


5 


1. 616 


.0250 


2. 761 


2.051 


all 


2! 


3ft 


ii 


ift 


4A 


2 


4* 


1 . 712 


.0277 


3.142 


2.302 


3i 


3A 


3f 


2 


ift 


4-!£ 


a* 


4i 


1 .962 


.0277 


3-976 


3-023 


3i 


3A 


4A 


2i 


2A 


4ll 


ai 


4 


2. 176 


.0312 


4.909 


3-7I9 


3* 


3ft 


4i 


ai 


2A 


5** 


2| 


4 


2.426 


.0312 


5 -94o 


4.620 


4i 


4A 


4§f 


2f 


2ft 


6 


3 


3* 


2.629 


•0357 


7.069 


5.428 


4f 


4A 


5l 


3 


2ft 


6ft 


3* 


3i 


2.879 


•0357 


8.296 


6.510 


5 


4tt 


5ft 


3i 


3A 


7A 


3i 


3i 


3.100 


•0384 


9.621 


7-548 


5§ 


5A 


6^ 


3i 


3A 


rH 


3i 


3 


3-317 


.0413 


11.045 


8.641 


5f 


5ft 


6ft 


3i 


3ft 


H 


4 


3, 


3-567 


.0413 


12.566 


9-963 


61 


6A 


7A 


4 


3ft 


m 


4i 


*£ 


3-798 


•0435 


14.186 


II.329 


6i 


6A 


7A 


4i 


4A 


9A 


4i 


a* 


4.028 


.0454 


15.904 


12-753 


6| 


6ft 


7§i 


4i 


4A 


9f 


4f 


af 


4.256 


.0476 


17. 721 


14.226 


71 


7A 


8ft 


4f 


4ft 


ioi 


5 


ai 


4.480 


.0500 


19-635 


I5-763 


7f 


7A 


8ft 


5 


4ft 


io|| 


5i 


2i 


4-73° 


.0500 


21.648 


17-572 


8 


m 


9A 


5i 


5A 


»H 


5* 


2| 


4-953 


.0526 


23-758 


19.267 


H 


»A 


9ft 


si 


5ft 


Hi 


si 


2f 


5-203 


.0526 


25.967 


21 .262 


H 


8ft 


10A 


5f 


5ft 


I2f 


6 


2i 


5-423 


•0555 


28.274 


23.098 


9k 


9A 


10ft 


6 


5ft 


I2ft 



142 



MACHINE DESIGN. 



The term " machine screws" covers many forms of small 

screws, usually with screw-driver heads. All of the kinds given 

in this classification are made in great variety of size, form, 
length, etc. 





Fig. 77. Fig. 78. 

Thus far American manufacturers have failed to agree upoD 
standard dimensions for set-screws and machine screws.* 

For consideration of their design, we will divide bolts and 
screws into three classes : 

(a) Those which are put under no stress by screwing up. 

(b) Those which are put under an initial stress by tightening. 

(c) Those which are used to transmit power. 

87. Analysis of Action of Screw.— Before taking these case* 
up in detail it will be well to examine into the general action 
of screw and nut. Reference is made to Fig. 79. The turning, 
of a nut loaded with W lbs. may be considered as equivalent 
to moving a load W on an inclined plane whose angle with 
the horizontal is the same as the mean pitch angle of the 
thread a. - ___ 

* The report of the committee of the A. S. M. E. on this subject with suggested 
standards, will be found in Vol. 28 of the Trans. A. S. M. E. 



BOLTS AND SCREIVS. 



143 



Let r\ = outside radius of thread ; 
r 2 = inside radius of thread; 

r = mean radius of thread, approximately -; 

P = pitch of thread; 

a=mean pitch angle, i.e., tan a 



2nr 



li = coefficient of friction between nut and thread ; 

<f> = angle of friction between nut and thread, i.e., tan <f> = /*. 

1st. To raise W. 

Consider W as a free body moving uniformly up the incline 
under the system of forces shown in Fig. 80, where W = axial 




T 

\ 

Fig. 80. 



load, R the normal reaction between nut and thread due to W, 
H the horizontal push forcing the nut up the incline, and F 
the friction in direction of incline due to the normal pressure R; 
then, by the ordinary laws of mechanics, 

R = W + cosa, (1) 

F=pW-r- cos a, (2) 

H = W tan (a + <f>) (3) 

The turning moment M 

= Hr = Wrtsin (a + <t>) (4) 



144 MACHINE DESIGN. 

P 
Since tan a = — , and tan d> = pt, while 

, . tan a + tan 6 

tan (a +</>) = : j, 

v r i— tan a tan <p 

it follows that 

M = Hr = Wr P + 27ir / (5) 

27ir — pjJL VJ/ 

2d. To lower W, 

H = W tan (a-4>) 9 (6) 

M = Hr = Wr t^ (7) 

The foregoing has applied to square threads. Consider V 
threads with /? = the angle of V with a plane normal to the axis 
of the screw. (See Fig. 81.) The mean helix angle = a as be- 




Fig. 81. 

fore, but now R slopes from W in two directions, making the 
angle a in the one, as before, but also making the angle /? with 
IF in a plane at right angles with the first. Hence 

R = W sec a sec/? (8) 

F = jiW sec a sec /? (9) 

For raising load, approximately, 

__ __ p + 27ir/xsec /? 

Hr = Wr- r- - Q (10) 



BOLTS AND SCREWS. 



145 



For lowering load, approximately, 

p — 2nrpt sec /? 



Hr = Wr 



27zr + pju sec /?* 



(«) 



88. Calculation for Screws Not Stressed in Screwing Up. — 

Returning now to (a). As illustrations of this class consider 
the eye-bolts shown in Fig. 82. It is customary to neglect the 
influence of the thread on the strength of the bolt, and to con- 
sider as the effective area, A, to resist stress, only the area of a 
circle whose diameter equals the diameter of the bolt at the 
bottom of the thread. In both cases considered a torsional 
stress is induced by screwing the engaging surfaces together, 
but if these surfaces are a proper fit, this stress is negligible, 
particularly since it exists only within the limits of the engaging 
threads where the action of the further working load which the 
bolt bears does not come into play. 





Fig. 82. 



The eye-bolt being now subjected to the working load T in 
the direction of its axis, we have 

A- v 

where / is the safe unit working stress for the material and con- 
ditions. 

If U = ultimate unit strength of the material, then, if the 



146 MACHINE DESIGN. 

load is a constant, dead load, / may be taken as great as — for 

o 
good wrought iron or mild steel. If the load is a variable one, 

slowly applied and removed, / should not exceed — for the 

same materials. If the load is variable and suddenly applied, 

/ should never exceed — for these materials, and in cases of 

shock may need to be much smaller than this. 

The case shown in Fig. 83 must not be confused with the 
preceding. Here (as will be explained under (b) ) there may be 
a tensile stress in A induced by compressing B-B between the 
shoulder and the nut. If the extension in the part A of the 
bolt due to the application of the force T later be greater than 
the compression caused in B-B by the tightening of the nut, 
then the shoulder will leave B-B, and simple tension = T results 
in all sections of the bolt below the nut, as in case (a). 

On the other hand, if the extension in the part A of the 
bolt due to the subsequent application of T be not so great as 
the original compression of B-B due to the screwing up, then 
we have in the part A a resultant tension greater than T. This 
case would come under (b). 

89. Calculation of Screws Stressed in Screwing Up. — (b) 
Combined tension and torsion are induced in a bolt by tighten- 
ing it. The stress may equal, or very greatly exceed, the tensile 
stress due to working forces. Consider the example shown in 
Fig. 78. Suppose the nut screwed down so that the parts con- 
nected by the bolt are held close together at E-F but have not 
yet been compressed. Suppose that the proportions are such 
that the wrench may be given another complete turn. The nut 
will move along the direction of axis of the bolt a distance = p. 
The parts held between the head and nut will be compressed 
and the body of the bolt will be extended. 



BOLTS AND SCREWS. 147 

The force applied at the point B, or end of the wrench 
(a distance / from the axis) will range from a value of o at the 
beginning of the turn to a value P at the finish. The average 

P 

value of the turning force will be approximately = -. 

The distance moved through by the point of application of 
this force is 2rd. Hence the work done in turning the nut a 
full turn under these conditions will be 



p 

- . 2rd = Pnl (12) 

2 



The resistances overcome by this application of energy are 
three in number: 

1st. The work done in extending the bolt. 

2d. The work done in overcoming the frictional resistance 
between nut and thread. 

3d. The work done in overcoming the frictional resistance 
between nut and washer. 

These will be considered in order. 

1st. Let T = the final pure tensile stress in the bolt due to 

screwing up one turn. At the beginning of the turn the ten- 

T 
sion = o. The average value may be considered = — for the 

turn. The distance moved through by the point of application 
of this force in the direction of its line of action, in one turn = ^. 
The work done in extending the bolt 

T 
=~P (13) 

2d. The frictional resistance between the threads of nut and 
bolt depends upon the form of the thread as well as the mate- 
rials used and the condition of the surfaces. (See equations (2) 



148 MACHINE DESIGN. 

and (9), § 87.) Assuming a V thread as being more commonly 
used for fastenings, the average value of the friction 

T 

F = ft— sec a sec /? 

T 

(eq. (9) ), since the average load for the turn=— . 

The distance moved through by the point of application of 
F for one turn of the nut on the bolt = ^> cosec a. Hence the 
work done in overcoming the friction between bolt and nut in 

T 

the one turn = jU— sec a sec /? . p cosec a 



T 
=— fip sec a sec ft . cosec a (14) 



3d. The frictional resistance between nut and washer due to 

T T 

a mean force — will be //— , in which u! is the coefficient of 

2 r 2 

friction between nut and washer. The point of application of 

3 
this resistance may be taken at a distance of —r\ from the axis 

of the bolt, r\ being the outside radius of bolt-thread. The dis- 
tance moved through by the point of application for one turn of 
the nut = 27rffi, and the work done in overcoming this frictional 
resistance 

T 
=-ti'27cn (15) 



Equating (12) to the sum of (13), (14), and (15), gives 



T T T 

Pnl=—p +—fip sec a sec /? cosec a-\—fi f $nr\ y 



BOLTS AND SCREWS. 



149 



Whence 



T = 



2Pnl 



p + ftp sec a sec /? cosec a + j/^7rri 



(16) 



ft =-7 = unit stress in bolt due to pure tension. . (17) 

In addition to this it must be borne in mind that the screw 
is subjected to a torsional moment whose value can be deter- 
mined by considering the nut as a free body as 
shown in plan view in Fig. 84, where all of the 
forces capable of producing moments about the 
axis of the nut are indicated as they exist at the 
end of the turn. 

Summing the moments about the axis of the 
bolt gives 

Fig. 84. Hr = Pl-p!T%n (18) 

Hr is, of course, the torsional moment transmitted from the nut 
to the bolt. To find its numerical value substitute the value of 
T found in equation (16) and solve (18). 

The unit stress induced in the outer fibers of a rod of cir- 
cular section and radius r 2 ( = radius at bottom of thread) is 
found by means of the equation 

J 




U 



M. 



(19) 



77T 2 

/ is the polar moment of inertia, in this case = ; c is the dis- 

r 2 

tance from neutral axis to most strained fiber, in this case =r% ; 
j 8 is the induced unit stress in outer fiber; M is the moment, in 
this case = i2>. Combining equations (18) and (19) and sub- 
stituting these values gives 

2{Pl-y!T\r,) 



h = 



7rr 2 '- 



(20) 



150 MACHINE DESIGN. 

The equivalent tensile unit stress due to the combined action 
of ft and f 8 is found from the equation for combined tension 
and torsion, 



; / = o.35/, + o.6 5 v7< 2 + 4/ 8 2 (21) 

90. Problem. — What is the unit fiber stress induced in a 
U. S. standard J-inch bolt in screwing up the nut with a pull of 
one pound at the end of a wrench 8 inches long ? Arrangement 
of parts as shown in Fig. 78. 

In this case d\ = .500 in., r\ = .25 in., 



d 2 = .400 in., r 2 = .2 in., 
^ = .225 in., A = .i26 sq. in., 
^ = .077 in., 
^=^=0.15, 

P 

a = angle whose tangent is — =3° f, 

sec a = 1.0015, cosec a = 18.39, 

= 3O°, sec £=1.155, 

P = ilb., and / = 8 ins. 

From equation (16) 

, = 2X1X^X8 _ 

"".077 + .077X0.15X1.0015X1. 155X18.39 + 0.15X3X^X0.25 
= 74.467 lbs. 

From equation (17), 



/( = 7fL467 = lbs . 

' .120 ° y 



From equation (20), 



t 2(1X8-0.15X74.467x1X0.25) 

7TX0.2 3 

=303 lbs." 



BOLTS AND SCREWS. 15 1 

From equation (21), 



7=0.35X591 +0-65 ' N/ 59i 2 + 4X303 2 
= 757 lbs. 

If a pull of one pound on an 8-inch wrench applied to a \- 
inch bolt can induce a unit fiber stress of 757 lbs., since equa- 
tions (16) and (20) show that the stress increases directly as the 
pull, it follows that a pull of 30 lbs., such as is readily exerted 
by a workman, will induce a stress of 30X757 = 22,710 lbs. per 
square inch. 

91. Wrench Pull. — If this turning up be gradual and the 
bolt is not subjected to working stresses, this would be safe for 
either wrought iron or mild steel. On the other hand, if the 
final turning be done suddenly by means of a jerking motion or 
a blow, or a long wrench be used, or even an extra-strong grad- 
ual pull be exerted, there is evident danger of / having a value 
beyond the elastic limit of the material, even reaching the ulti- 
mate strength. 

It will be noticed also that the torsional action increases the 

7^7 
fiber stress over that due to pure tension in the ratio of — , i.e., 

59 1 

in this problem, an increase of over 25 per cent. In general this 

increase will be from 15 to 20 per cent, depending chiefly upon 

the relation existing between jj. and //. It should also be noted 

that the pure tension, T, induced in the bolt by the moment PI 

may be taken as the measure of the pressure existing between 

the surfaces E-F (Fig. 78). In our problem this pressure, for 

P=3o lbs., would become 30X74.467=2234 lbs. 

As a general rule the length of wrench used by the workman 

is fifteen or sixteen times d i} the diameter of bolt, and it may be 

stated that T = j$P for U. S. standard threads. 



152 



MACHINE DESIGN. 



92. Calculation of Bolts Subject to Elongation.— Next con- 
sider the case shown in Fig. 85. Suppose that the nut is screwed 



up with a resulting tensile stress in the bolt 
= T. A working force Q tends to separate 
the bodies A and B at C-D. Assume that Q 
acts axially along the bolt. The question is, 
What value may Q have without opening the 
joint C-D? 

A is the cross-sectional area of the bolt; 

L is the original length between bolt-head 
and nut when A and B are just in 
contact at C-D but not compressed ; 

To is the tensile stress in bolt due to sere wing up ; 

X is the total elongation of bolt due to To ; 

E is the coefficient of elasticity of the bolt material. 

r™ • un i t strain 1 . . „ 

I hen, since — r—- =-5, it follows that 



r rp. 




Fig. 85. 



unit stress E' 



L_ 
A 



(1) 



In any given case this can be solved for X. 
Let A' be the area of A and B compressed by the bolt action; 
X' is the total compression (i.e., shortening) of A and B, 

due to the tightening of the bolt ; 
Co is the total compressive stress which produces X f \ 

.'. C = To. 
E r is the coefficient of elasticity of the material A, B. Then 

L_ _i_ 

Co~E r ' ° e * ° " ' 
A' 



(2) 



BOLTS AND SCREWS. 1 53 

This can be solved for A' . 

Now consider the condition when a working force, Q, acts 
tending to elongate the bolt so that A and B will just be ready 
to separate at C-D. In order that this separation may begin, 
the bolt, already elongated an amount A, must be elongated a 
further amount A' . 

For incipient separation the total elongation of the bolt then 
=A+A', and the total stress in the bolt corresponding to this 
elongation, = T' } can be determined from the equation 



l+X' 




L 
V 


i 
= E 



(3) 



H< 



Considering the bolt-head as a free body (Fig. 86), it follows 
' Y1 that the forces acting on it at any instant will be, C, 
t the reaction of the material of A due to its resistance 
j to compression; Q, the working force; and T, the ten- 
Fig. 86. sion in the bolt. Hence 

T = C + Q ( 4 ) 

When the bolt is first screwed up, and Q=o, then C = T y 
and T = T , the tension due to screwing up. When Q comes 
into action, C is partly relieved, and when Q reaches such a 
value that the surfaces are about to part, then C=o and Q = 
T = T. (See equation (3).) 

An examination of these formulae shows certain facts which 
may be stated as follows: The tightness of the joint C-D de- 
pends upon the compressibility of A and B. 

Anything which increases the total compression, A' f increases 
the tightness of the joint. This may be accomplished by in- 
creasing L or Co, or decreasing A' . It may also be increased 



154 



MACHINE DESIGN. 



by the introduction of a highly elastic body (i.e., gasket) between 
A and B. 

It also follows that the tension in the bolt when the joint is 
about to open, T', must be greater than the tension due to 
screwing up, T , and therefore if Q be limited to a value equal 
to or less than T , there will be no opening of the joint. In 
general, A' is large compared with A, and \' very small com- 
pared with X, so that T f is not much greater than To. In order 
to be sure of a tight joint the initial tension should be taken 

To = 2Q. 

93. Problem I. — Calculate the bolts for a " blank' ' end for 
a 6-inch pipe using flanged couplings with ground joints, and 
no gaskets, as shown in Fig. 87. The excess internal pressure 
is to be 150 lbs. per square inch. 




Fig. 87. 
The area subjected to pressure has a diameter of 7^ ins.; 



hence the total working pressure = 150 X 



*X7-5' 



6627 lbs. 



The number of bolts is determined by the distance they may 
be spaced apart without danger of leakage due to the springing 
of the flange between the bolts. This distance may be taken 
equal to four or five times the thickness of the flange. In the 
problem under consideration, the diameter of the bolt circle will 
be approximately 9 ins., and using six bolts, the chord length 
between consecutive ones will be about \\ ins., which is per- 
fectly safe. 



With six bolts 



BOLTS AND SCREWS. 155 

^ 6627 

Q -=-* =1105 lbs. 



Take To =2<2 = 2210 lbs. 

With a direct tension of 2210 lbs. due to screwing up, there 
is also the stress due to torsion. As stated in § 91, this may- 
increase the fiber stress 20 per cent over that due to direct ten- 
sion. To allow for this the bolts used must be capable of safely 
sustaining a stress of 2210X1.20 = 2650 lbs. 

The allowable unit stress here may be taken rather high, 
since the conditions after once screwing up approximate a steady 
load. Assume steel bolts with an allowable unit stress of 
15,000 lbs. 

The area of each bolt at the bottom of the thread will then 

2650 

be =0.177 sq. in. This value lies between a i^-inch and 

15000 " ^ 

a f-inch bolt. Select the latter with an area of 0.202 sq. in. To 
exert an initial tension of 2250 lbs. in a f-inch bolt would re- 
quire a pull of about 30 lbs. on a 10-inch wrench. (See § 91.) 
These values just about correspond to actual conditions in prac- 
tice. 

94. Problem II. — It is required to design the fastenings to 
hold on the steam-chest cover of a steam-engine. The opening 
to be covered is rectangular, io"Xi2". The maximum steam- 
pressure is 100 lbs. per square inch. The joint must be held 
steam-tight. Studs of machinery steel having an ultimate ten- 
sile strength of 60,000 lbs. per square inch will be used. 

The total working pressure = 10 X 12 X 100 = 12,000 lbs. 

The number of studs to be used will be governed by the dis- 
tance they may be spaced without springing of the cover. The 
thickness of the latter being assumed to be f inch at the edge, 
the spacing should not exceed 5 Xf" = y, say 4 inches. 



*56 



MACHINE DESIGN. 



c 



c 



c 



c 



£L 



n. 



-10- 



The opening is io"X 12", as shown in Fig. 88. There must 
be a band about f or f inch wide 
around this for making the joint, 
upon which the studs must not en- 
croach. This makes the distance 
between the centers of the vertical 
rows of studs about 14 inches, and 
between the horizontal rows about 
12 inches. Twelve studs can be used 
if arranged as shown in the figure. 
The greatest distance, that between 7 r 

the studs across the corners, will but fig. 88. 

slightly exceed the allowable 4 inches. 







3 



3 



3 



With 12 studs, the working load on each = Q 
T = 2Q = 2000 lbs. 



12000 
12 



1000 lbs. 



Allowing 20 per cent for torsional stress, increases this to 

2400 lbs. 

Allowing a unit stress of 15,000 lbs., as in Problem I, we 

2400 

have as the area of the stud at the bottom of thread = 0.160. 

15000 

This corresponds to a ^-inch stud. Since a workman may 
readily stress a bolt of this size beyond the elastic limit by exert- 
ing too great a pull in tightening, many designers would increase 
these studs to f inch or even f inch. 

95. Design of Bolts for Shock. — The elongation of a bolt 
with a given total stress depends upon the length and area 
of its least cross-section. Suppose, to illustrate, that the bolt, 
Fig. 89, has a reduced section over a length / as shown. This 
portion, A, has less cross-sectional area than the rest of the bolt, 
and when any tensile force is applied, the resulting unit stress 
will be greater in A than elsewhere. The unit strain, or elonga- 
tion, will be proportionately greater up to the elastic limit; and 



BOLTS AND SCREWS. 



157 




if the elastic limit is exceeded in the portion A, the elongation 
there will be far greater than elsewhere. If there is much differ- 
ence of area and the bolt is tested to rup- 
ture, the elongation will be chiefly at A. 
There would be a certain elongation per 
inch of A at rupture. Hence the greater 
the length of A, the greater the total elonga- 
tion of the bolt. If the bolt had not been 
reduced at A, the minimum section would 
be at the root of the screw-threads. The 
Fig. 89. Fig. 90. ax ial length of this section is very small. 
Hence the elongation at rupture would be small. Suppose there 
are two bolts, A with and B without the reduced section. They 
are alike in other respects. They are subjected to equal tensile 
shocks. Let the energy of the shock = E. This energy is di- 
vided into force and space factors by the resistance of the bolts. 
The space factor equals the elongation of the bolt. This is 
greater in A than in B, because of the yielding of the reduced 
section. But the product of force and space factors is the same 
in both bolts, =E; hence the resulting stress in the minimum 
section is less for A than for B. The stress in A may be less 
than the breaking stress, while the greater stress in B may 
break it. The capacity of the bolt to resist shock is 

THEREFORE INCREASED BY LENGTHENING ITS MINIMUM SEC- 
TION TO INCREASE THE YIELDING AND REDUCE STRESS. This is 

not only true of bolts, but of all stress members in machines. 

The whole body of the bolt might have been reduced, as 
shown by the dotted lines in Fig. 89, with resulting increase of 
capacity to resist shock. Turning down a bolt, however, weak- 
ens it to resist torsion and flexure, because it takes off the material 
which is most effective in producing large polar and rectangular 
moments of inertia of cross-section. If the cross-sectional area 
is reduced by drilling a hole, as shown in Fig. 90, the torsional 



158 



MACHINE DESIGN. 



and transverse strength is but slightly decreased, but the elon- 
gation will be as great with the same area as if the area had 
been reduced by turning down. 

Professor Sweet had a set of bolts prepared for special test. 
The bolts were i\ inches diameter and about 12 inches long. 
They were made of high-grade wrought iron, and were dupli- 
cates of the bolts used at the crank end of the connecting-rod 
of one of the standard sizes of the Straight-line Engine. Half 
of the bolts were left solid, while the other half were carefully 
drilled to give them uniform cross-sectional area throughout. 
The tests were made under the direction of Professor Carpenter 
at the Sibley College Laboratory. One pair of bolts was tested 
to rupture by tensile force gradually applied. The undrilled 
tolt broke in the thread with a total elongation of 0.25 inch. 
The drilled bolt broke between the thread and the bolt-head 
with a total elongation of 2.25 inches. If it be assumed that 
the mean force applied was the same in both cases, it follows 
that the total resilience of the drilled bolt was nine times as great 
as that of the solid one. "Drop tests," i.e., tests which brought 
tensile shock to bear upon the bolts, were made on other similar 
pairs of bolts, which tended to confirm the general conclusion. 

96. Problem. — It is required to design proper fastenings for 
holding on the cap of a connecting-rod like that shown in Fig. 
91. These fastenings are required to sus- 
tain shocks, and may be subjected to a 
maximum accidental stress of 20,000 lbs. 
There are two fastenings, and therefore 
each must be capable of sustaining safely 
a stress of 10,000 lbs. They should be ' m i k 
designed to yield as much as is consistent 
with strength ; in other words, they should 
be tensile springs to cushion shocks, and 
thereby reduce the resulting force they have to sustain. Bolts 








{□J 



Fig. 



91. 



BOLTS AND SCREWS. 159 

should therefore be used, and the weakest section should be 
made as long as possible. Wrought iron will be used whose 
tensile strength is 50,000 lbs. per square inch. The stress given 
is the maximum accidental stress, and is four times the working 
stress. It is not, therefore, necessary to give the bolts great 
excess of strength over that necessary to resist actual rupture 
by the accidental force. Let the factor of safety be 2. This 
will keep the maximum fiber stress within the elastic limit. 
Then the cross-sectional area of each bolt must be such that 
it will just sustain 10,000X2=20,000 lbs. This area is equal 
to 20,000 -=-50,000 =0.4 sq. in. This area corresponds to a 
diameter of 0.71 inch, and that is nearly the diameter of a 
J-inch bolt at the bottom of the thread ; hence J-inch bolts will 
be used. The cross-sectional area of the body of the bolt 
must now be made at least as small as that at the bottom of the 
thread. This may be accomplished by drilling. 

97. Jam-nuts. — When bolts are subjected to constant vibra- 
tion there is a tendency for the nuts to loosen. There are many 
ways to prevent this, but the most common one is by the use of 
jam-nuts. Two nuts are screwed on the bolt; the under one 
is set up against the surface of the part to be held in place, and 
then while this nut is held with a wrench the other nut is screwed 
up against it tightly. Suppose that the bolt has its axis vertical 
and that the nuts are screwed on the upper end. The nuts being 
screwed against each other, the upper one has its internal screw 
surfaces forced against the under screw surfaces of the bolt, and 
if there is any lost motion, as there almost always is, there will 
be no contact between the upper surfaces of the screw on the 
bolt and the threads of the nut. Just the reverse is true of the 
under nut; i.e., there is no contact between the under surfaces 
of the threads on the bolt and the threads on the nut. There- 
fore no pressure that comes from the under side of the under 
nut can be communicated to the bolt through the under nut 



i6o 



MACHINE DESIGN. 



directly, but it must be received by the upper nut and com- 
municated by it to the bolt, since it is the upper nut alone that 
has contact with the under surfaces of the thread. Therefore 
the jam-nut, which is usually made about half as thick as the 
other, should always be put on next to the surface of the piece 
to be held in place. Other locking devices are shown in Fig. giA. 





CT 





i^^S 



t^j- 



Fig. 91^4. 

98. Calculation of Screws for Transmission of Power. — 

(c) Screws are frequently used to transmit power. A screw-press is 
a common example, while the action of spiral gears, including worms 
and wheels, is that of screws and subject to the same analysis. 
Collar friction, or nut and washer friction, is here neglected. 
The use of ball or roller thrust bearings permits this. Equation 
(18) shows how it may be introduced if PI does not equal Hr. 

The general action of screw and nut has already been treated. 
(See § 87.) 

With a square-thread screw it was found that the moment 

<b -(- 27TTU. 

Pl=M, required to raise a load W, will be=Wr — (z) 

' ^ 27rr — pjj. VJ 



(page 144). 



c(PD 



This will induce a fiber stress j s = — j— (see equation (19) ), 



BOLTS AND SCREIVS. 



161 



W 

which must be combined with the tension, } t =~r, in order to 

get the actual unit stress, /, remembering 



/ = .35/< + -65^ 2 +4/* 2 (21) 

Let n= number of complete thread surfaces in contact 
between the nut and screw, and the projected area equals 

n- (di 2 — d 2 2 ) to bear the load W. 
4 

W=Kn-(d 1 2 -d 2 2 ), 
4 

where K is the allowable pressure per square inch of projected 
thread area. 

For nuts and bolts which are used as fastenings we may take : 
Z-2500 lbs. for wrought or cast iron running on the same 

material or on bronze ; 
K = 3000 lbs. for steel on steel or bronze. 

With good lubrication, where the screw and nut are used to 
transmit power, we may take the values given in the following 
table : 

Table VII. 



Rubbing Speed in Feet 
per Minute. 



50 or less. 
100 

150 

250 

400 



Value of K. 



Iron. 



Steel. 



2500+ 


3000+ 


1250 


1500 


850 


IOOO 


400 


500 


200 


250 



The value of /1 has been experimentally determined by Pro- 
fessor Kingsbury.* He concludes that for metallic screws turn- 



* Transactions A. S. M. E., Vol. XVII, pp. 96-116. 



162 



MACHINE DESIGN. 



ing at extremely slow speeds, under any pressure up to 14,000 
lbs. per square inch of bearing surface, and freely lubricated 
before application of the pressure, the following coefficients of 
friction may be used. 



Table VIII. 



Lubricant. 


>t. 




0. 11 
0.143 

0.07 




Heavy-machinery oil and graphite in equal 





Regarding the efficiency of the square-screw thread to trans- 
mit energy, we may reason as follows : 

_, rr . useful work 

The efficiency = ,= totalwork , 



which becomes for one turn = 



Wp 



Wp 2nr 



27zrH' 
W tan a 



tan a 



2nrH H IF tan (<* + <£) tan (« + <£)' 



(22) 



From this it appears that e becomes o, for a=o° and for 
a = 90 — <£, and must therefore have a maximum value between 
these limits. To determine this maximum, write 



tan a 



= tan a cot (a + $) . 



tan (a + 4>) 

Taking the first differential and equating to o, 

tan a 



cot (a + <f) 
c 

Solving which gives a== 45" 



cos 2 a sin 2 (a + <t>) 

o 4> 



= 0. 



BOLTS AND SCREWS. 163 

To find the corresponding value of e, write (from (22) ) 



tan 
max. e 



(«•-*) t an( 4S °-J) 



tan (45°-^-+^ tan( 4 5 +D 



To lower PF with a square-threaded screw, 

P/ = ia> = W ^-Z|^ ( ) 

[The value of j 8 can be found from this, as explained in the 
section on raising W, and combined with f t to obtain /.] 
Regarding the efficiency in this case, 

if a< <j> the load will not sink of itself (i.e., overhaul), 

if a = <f> we have a condition of equilibrium, 

if a ></> the load will sink of itself (i.e., overhaul). 

For a screw which will not overhaul it becomes evident that 
the limiting value of a is <f> and the maximum efficiency 

tan a tan </> 1 —tan 2 <j> 

e = - 7 — —rt =- 7 = =o.«5— o.«; tan 2 <b. 

tan (a + 0; tan 29 2 j j 7- 

The efficiency of a screw which will not overhaul can there- 
fore never exceed 0.5 or 50 per cent. 

For V threads, with ,5 = angle of V, with a plane normal to 
the axis of the screw for raising load, 

n-Wf > + ~J""l ( 24 > 

2nr — pn sec p v 

which is evidently greater than (5), and 

tan a(i — p tan a sec B) x 

e= : ; a 1 .... (25; 

tan a 4 p sec /? v J 



1 64 MACHINE DESIGN. 

which is evidently less than e for square threads. (See equa- 
tion (22).) 

It is clear, then, that square threads should be used in pref- 
erence to V threads for screws for power transmission. 

For lowering the load with V threads 

P;-gr-H ^-?7 sec ; (26) 

2nr + pfi sec p K 

99. Problem. — Design a screw to raise 20,000 lbs. The 
screw must not overhaul. No collar friction. 

What moment need be exerted to lift the load ? 

What will be the efficiency of the screw ? 

Select a square-thread screw of machinery steel running in 
a bronze nut. 

For a screw which will not overhaul a must be less than <j>. 

:. tan a< /i. 

To be safe against overhauling with the materials used and 
good lubrication, /* must not be given a greater value than 0.10. 

.-. a<5°45' and = 5°45'- 

The pure tension = 20,000 lbs. = W. In the preliminary calcu- 
lations, to allow for the effect of torsion, this will be increased 

. 25000 
so that 7 = — -: — . 

' A 

In this equation / is the allowable unit stress in pounds per 
square inch, and A is the area of the screw at the bottom of the 
thread in square inches. 

Assume that this screw is frequently loaded and unloaded, 

and not subject to shocks nor reversal of stress so that / = 

25000 

12,000 lbs. per square inch for mild steel. Then A = = 

r ^ 12000 



BOLTS AND SCREWS 165 

2.08 sq. ins. This corresponds to a diameter of if inches at 
bottom of thread. 

P 

From tan a = , we nave 

27rr 

p = 27ir tan a. 
Remembering that for square threads the depth of the thread 

= -£, and that r 2 is the radius at the bottom of thread, and 
2 

:. r = r 2 + — , it follows that 



p = 2nl 



r 2 + - ) tan a, 



27t tan a 

p p = 27tr 2 tan a, 

4 



27zr 2 tan a 

*-— — < 

i—— tan a 

2 



2X71X0.8125X0.1 . . . , 

.". # = = .000 inch. 

r 1 — 1.57X0.1 

This is not a thread to be easily cut in the lathe. It would 
be desirable to modify the value of p so that the thread can be 
readily cut. It is obvious that p cannot be increased without 
increasing r proportionately, else a will have a greater value 
than is allowable. It will be more economical to reduce p. 
The nearest even value would be \ inch, and this will be selected. 
Check this for strength: 
^2 = 1-625 inches, r 2 =0.8125 inch, ^ = 0.5 inch. 

d\ = 2.125 inches, r\ = 1.0625 inches, 
d = 1.875 inches, r =0.9375 inch, 



i66 MACHINE DESIGN. 

tan a:=— = Ky °' 5 = 0.087, 

2nr 2X^X0.9375 

which is safe, as it is less than the value of // = o.io. 
From equation (5), the moment, 

Pl=Hr = Wr p±^i. 

27ir — pfi 

_, w 0.5 + 2X^X0.9375X0.10 

,\ Pl = 200OOX0.9375 , J vy x , J w r 

yo/:) (2X7rXo.9375) -(0.5X0.10) 

= 3496 in. -lbs. 
From equation (19), the fiber stress due to torsion, 
cPl 2PI 



2X349 6 
7rXo.8i25 i 



/a= ^ rN/ ^ QT ^ rj3 = 4i4Q lbs. per square inch. 



From equation (17), the unit stress due to tension, 

. W 20000 „ . . 

ft=-j- = = 9597 lbs. per square inch. 

From equation (21), the combined stress, 



/ = o.35/, + o.6 5 V7, 2 + 4 / fi 2 . 



•'• /=o.35X9597+o.65V9597 2 +4X4i4o 2 
= 12,379 lbs. per square inch, 

which is near enough 12,000 to be considered safe. 
The efficiency, from equation (22), 



tan a 
e = 



tan (a + cp) 



BOLTS AND SCREWS. 167 

Since tan a = 0.087, a = 5 . 
Since ju. = tan <f> = o. 10, $ = 5 45'. 

,\ « + </> = io° 45', tan io° 45' = 0.1899. 

••• e= ^ =o - 458I ' 0r45 - 8I% - 

The height of the nut is determined from the equation 

W = Kn-(d 1 2 -d 2 2), 
4 

in which W is the load, n the number of complete threads in 
the nut, d\ the outside and d 2 the inside diameter of thread. 
K is the allowable pressure in pounds per square inch, and its 
value depends upon the speed. See table in sec. 98. 

Assuming the screw to have a rubbing velocity of less than 
50 feet per minute, K= 7,000. Then 

W 20000 

n=- 



*,! o j 2x 3 00 °Xo.7854(2.i25 2 -i.625 2 ) 



3000- (di 2 — d 2 2 ) 
4 

= 4.5, nearly. 
The height of nut = ^Xw = J"X4.5" = 2£", 



CHAPTER IX. 

MEANS FOR PREVENTING RELATIVE ROTATION. 

ioo. Classification of Keys. — Keys are chiefly used to pre- 
vent relative rotation between shafts and the pulleys, gears, etc., 
which they support. Keys may be divided into parallel keys, 
taper keys, disk keys, and feathers or splines. 

101. Parallel Keys. — For a parallel key the "seat," both 
in the shaft and the attached part, has parallel sides, and the 
key simply prevents relative rotary motion. Motion parallel to 
the axis of the shaft must be prevented by some other means, 
as by set-screws which bear upon the top surface of the key, as 
shown in Fig. 92. A parallel key should fit accurately on the 
sides and loosely at the top and bottom. Parallel keys may be 
" square " or " flat." The following table (IX) for dimensions 
for square keys is from Richards's " Manual of Machine Con- 
struction." 

Table IX. 
Diameter of shaft = d= 1 1 \ 



1 
2 


If 


2 


2i 


3 


3i 


9 
32 


H 


ft 


M 


17 
32 


& 


16 


3 
8 


A 


1 
2 


£ 


5 

8 



Width of key = w= -^ *2 

Height of key = /= ^ J A. I ^ \ At ! 

Excellent parallel keys are made from cold-rolled steel with- 
out need of any machining. 

d 

John Richards's rule for flat keys is (see Fig. 93) w = —. /has 

such value that a = 30.° This rule is deviated from somewhat, 

as shown by the following table (X), taken from his " Manual 

of Machine Construction," page 58: 

168 



MEANS FOR PREVENTING RELATIVE ROTATION. 169 














Table X. 












1 

4 


I* 


If 


2 


a* 


3 3§ 


4 


5 


6 


7 


8 


5 

T6 


f 


7 
16 


1 
2 


5 

8 


3 7 

4 8 


1 


il 


if 


if 


1 


3 
16 


1 
4 


9 
32 


5 
16 


3 

8 


A i 


5 

8 


11 

16 


13 
16 


7 

8 


1 



When two or more keys are used, w=d~6 J t being, as before, 
of such value that a shall =30°. 

102. Taper Keys. — A taper key has parallel sides and has 
its top and bottom surfaces tapered, and is made to fit on all 



H_ 




Fig. 92. Fig. 93. Fig. 94. 

four surfaces, being driven tightly " home." It prevents rela- 
tive motion of any kind between the parts connected. If a key 
of this kind has a head, as shown in Fig. 94, it is called a " draw 
key," because it is drawn out when necessary by driving a 
wedge between the hub of the attached part and the head of 
the key. Projecting draw-heads are to be avoided on all rotating 
parts unless guarded to prevent accident. When a taper key 
has no head it is removed by driving against the point with a 
" key-drift." 

The taper of keys varies from J to J inch to the foot. 

103. Fitting Shaft and Hub. — In using taper keys it is cus- 
tomary to bore out the hub slightly larger than the diameter of 
the shaft so that the wheel may be removed readily after the 
key is withdrawn. This allowance in diameter should not be 
greater than that for a sliding fit, say, 

0.05^+0^ 



iOOO 



170 



MACHINE DESIGN. 



in which formula A is the difference in diameter between the 
bore of hub and size of shaft, expressed in decimal parts of an 
inch, and D is the nominal diameter of shaft in inches. Where 
the parts do not have, to be taken apart frequently, it is vastly 
better to use a driving fit, i.e., to bore out the hub smaller than 

D 



the diameter of the shaft by an amount A 



iooo 



and to use 



parallel keys. 

Where a single taper key is used the effect is to make the 
wheel and shaft eccentric, as can be seen in Fig. 95. The bear- 
ing is limited to two points, A, B, and the connection is unstable 
for the transmission of power. 






Fig. 95. 



Fig. 96. 



Fig. 97. 



If great care is not exercised in having the taper of keyway 
exactly the same as the taper of the keys, a further difficulty 
arises in that the wheel will be canted out of a true normal plane 
to the shaft-axis. This can be seen in Fig. 96. 

By using two keys, placed a quarter or third of the circum- 
ference apart, a much more stable connection is obtained, as it 
will give three points of bearing, A, B, and C. {See Fig. 97.) 
Eccentricity is not avoided by this method. 

104. Woodruff Keys. — The Woodruff or disk system of keys 
is used by some manufacturers. The key is a half disk, as can 



MEANS FOR PREVENTING RELATIVE ROTATION. 



171 



be seen in Fig. 98. Under this system the keyway is cut lon- 
gitudinally in the shaft by means of a milling-cutter. This cut- 





Fig. 98. 



Fig. 99. 



ter corresponds in thickness to the key to be inserted, and is of 
a diameter corresponding to the length of the key. The key 




E 



Fig. 100. 



being semicircular, it is sunk into the shaft as far as will allow 
sufficient projection of the key above the surface to engage the 
keyway in the hub. 



172 MACHINE DESIGN. 

Owing to its peculiar shape the key may be slightly inclined, 
so that it will serve to support the wheel on a vertical shaft, pro- 
vided the key-seat in the hub is made tapering and of the proper 
depth. 

104. Saddle, Flat and Angle Keys. — Saddle keys (Fig. 99, A) 
and keys on flats (Fig. 99, B) are used occasionally. They 
have not the holding power of sunk keys. 

The type of key shown in Fig. 100 A has much to be said in 
its favor both as regards ease and accuracy in obtaining a stable 
connection and also as regards suitability of form to resist stresses. 
It will be noted that the surfaces are normal to the lines of action 
of the forces transmitted. The pressure per square inch should 
not exceed 17,000 lbs. The height of key is taken equal to 
0.2 diameter of shaft. 

The Kernoul key, shown in Fig. 10c B, is for use in driving 
in only one direction. A portion of the hub is cut out so as to 
form an eccentric slot. In this the key fits as shown. The 
inner face of the key, curved to the radius of the shaft, should 
be left rough so as to seize the shaft, while the outer face, curved 
to fit the slot in the hub, is smooth finished. When the shaft 
rotates (in this case) counter-clockwise, the resistance to the 
hub's motion being then as indicated by the arrow, the surface 
of the slot tends to slide up on the key, causing it to wedge in 
between the shaft and hub, forming a firm connection. When 
the shaft rotates in the opposite direction and the resistance 
to the hub's motion is reversed, the slot of the hub tends to leave 
the key, relieving the pressure and permitting easy removal from 
the shaft. At " a " and "b " there are counter-sunk screws for 
setting up and loosening the key. In Fig. 100 C is shown a special 
form of this type of key. It is known as the Barbour key and 
is chiefly used for fastening the cams on the shafts of stamp 
mills.* 

* Patent held for this purpose by the Risdon Iron Works. The distinguishing 
feature from the plain Kernoul key lies in the use of the inside projecting tongue 
which fits in a key way cut in the shaft 



MEANS FOR PREVENTING RELATIVE ROTATION. 173 

In the study of keys which drive in one direction only it is 
proper to include the roller ratchet. The simplest form is shown 
in Fig. 100 D. The hub is recessed as shown and the roller R 
placed in the recess, held in position by a spring. The direction 
of shaft rotation and hub resistance being as shown the roller 
becomes wedged between the two, forming a driving connection. 
With reversal of direction the roller is freed and the shaft and 
hub may have relative motion. 

Generally more than one roller is used and the mechanism 
takes the form shown in Fig. 100 E. A is a hardened and 
ground steel ring or bushing and B should also be hardened. 
Each roller should be held in place by a spring as shown at C. 
Such ratchets permit of rapid reciprocation. Complete details 
and descriptions of further clutches of this type will be found in 
the American Machinist of Dec. 21, 1905. 

105. Strength of Sunk Keys. — The strength of the latter is 
the measure of their holding power. A key of width =w, thick- 
ness =/, length =/, unit shearing strength =/ s , and unit crushing 
resistance =/ c will have a shearing strength - j s wl and a crush- 
ing resistance ) c \tl. 

If r= radius of the shaft, the moment which the key can 
resist will be measured by rwl) s or \rtl] c , whichever is smaller 
in value. 

All dimensions being expressed in inches and resistance 
in pounds per square inch, the moments will, of course, be 
expressed in inch-pounds. Experiments made by Professor 
Lanza indicate that the ultimate value for } s for cast iron = 
30,000 lbs., for wrought iron = 40,000 lbs., and for machinery 
steel = 60,000 lbs. A factor of safety of at least 2 would b: 
advisable with these values. 

H. F. Moore has experimented on the effect of key-ways 
on strength of shafts.* He deduces the following expressions 
for effect on stiffness and strength, respectively: 

* Bulletin No. 42, Univ. of Illinois, Eng. Exp. Station. 



1 74 MACHINE DESIGN, 

k = 1.0+0.40/ +0.7&, 

e= 1.0 -o.2«/' — 1.1A; 

£ = ratio of angle of twist of shaft with key-way to angle of 

twist of similar shaft without key-way. 
e = ratio of strength at elastic limit of shaft with key- way to 
the strength at elastic limit of a similar shaft without 
key-way; called efficiency; 
w' = width of key-way 4- diameter of shaft; 
h = depth' of key-way ~ diameter of shaft. 
The equations hold for values of w' up to 0.5, and for h up 
to 0.1875. The efficiency was not affected by the addition of a 
bending moment as great as the twisting moment. The efficiency 
of a shaft with two key-ways cut in the same plane for two 
Woodruff keys, of such size that the strength of the solid shaft 
was equal to the shearing strength of the two Woodruff keys, 
was found to be about the same as the efficiency of a shaft 
with a key-way whose width was one-fourth and whose depth 
was one-eighth of the shaft diameter, being about 85 per cent 
in both cases. 

106. Feathers or Splines are keys that prevent relative rota- 
tion, but purposely allow axial motion. They are sometimes 



Fig. ioi. Fig. 102. 

made fast in the shaft, as in Fig. 101, and there is a key "way" 
in the attached part that slides along the shaft. Sometimes the 
feather is fastened in the hub of the attached part, as shown 
in Fig. 102, and slides in a long key way in the shaft. 

It is frequently undesirable to have the feather loose. In 
such cases it is common to use tit-keys as shown in Fig. 103. 



MEANS FOR PRESENTING RELATIVE ROTATION. 



175 



The keys may be fastened to either hub or shaft. The tits 
are forged on the keys. Corresponding holes are drilled 
and countersunk in the piece to which the key is to be fastened, 
and after the key is placed in position the ends of the tits are 
riveted over to hold it securely in place. 

Machine screws are sometimes used in place of tits, but 
they suffer from the disadvantage of jarring loose. 

A satisfactory way of holding a key in a hub is shown in 
Pig. 104. 

Where the end of a stud is to receive change-gears a con- 
venient form of key is the dovetail shown in Fig. 105 in cross- 




Fig. 103. 



Fig. 104. 



Fig. 105. 



section. The dovetailed key-seat is generally cut with a mill- 
ing-cutter, and is made a tight fit for the key. After the latter 
is in place the shaft is calked against it at A- A. 
For feathers, Richards gives: 









Table XI 


IV = \ 
*= 1 


1* 
J 

1 


if 
A 


2 2\ 2 

A t 



\ 



I 



4* 
f 

7 



107. Round Taper Keys. — For keying hand-wheels and 
other parts that are not subjected to very great stress, a cheap 
and satisfactory method is to use a round taper key driven 
into a hole drilled in the joint, as in Fig. io6^4. If the two parts 
are of different material, one much harder than the other, 



176 



MACHINE DESIGN. 



this method should not be used, as it is almost impossible in 
such case to make the drill follow the joint. For these keys 
it is customary to use Morse standard tapers, as reamers are 
then readily obtainable. 

108. A Cotter is a key that is used to attach parts that are 
subjected to a force of tension or compression tending to sepa- 




Fig. 106. 

rate them. Thus piston-rods are often connected to both pis- 
ton and cross-head in this way. Also the sections of long 
pump-rods, etc. Fig. 106, B, shows a taper pin cotter. 

Fig. 107 shows machine parts held against tension by cot- 
ters. It is seen that the joint may yield by shearing the cot- 




Fig. 107. 

ter at AB and CD) or by shearing CPQ and ARS; by shear- 
ing on the surfaces MO and LN; or by tensile rupture of the 
rod on a horizontal section at LM. All of these sections should 
be sufficiently large to resist the maximum stress safely. The 
difficulty is usually to get LM strong enough in tension; but 
this may usually be accomplished by making the rod larger 



MEANS FOR PRESENTING RELATIVE ROTATION. 



177 



or the cotter thinner and wider. It is found that taper sur- 
faces if they be smooth and somewhat oily will just cease to 
stick together when the taper equals 1.5 inches per foot. The 
taper of the rod in Fig. 107 should be about this value in order 
that it may be removed conveniently when necessary 

From consideration of the laws of friction it is obvious that 
where a taper cotter is used, either alone as in Fig. 108 or in 
connection with a gib as in Fig. 109, the angle of taper a must 
not exceed the friction angle <£. That is, if the coefficient of 
friction be /*, then /* = tan <j> and tan a must be less than tan <j> 
or /1. Since, for oily metallic surfaces, pt may have a value 



r — <s^| 





Fig. 108. 



Fig. 109. 



as low as 0.08, it follows that a must not exceed 4^°. If both 
surfaces of the cotter slope with reference to the line of action 
of the force, the total angle of the sloping sides must not 
exceed 9 . 

109. Set-screws (see § 86, p. 140) are frequently used to pre- 
vent relative motion. They are inadvisable for heavy duty 
and if used on rotating members never should have projecting 
heads. Experiments made by Professor Lanza * with f -inch 
wrought-iron set-screws, ten threads to the inch and tightened 
with a pull of 75 lbs. at the end of a 10-inch wrench, gave the 
following results : 



* Trans. A. S. M. E., Vol. X. 



17^ MACHINE DESIGN. 



Table XII. 



■rr- ■< f tj„. Holding Power at 

Kind of End. Surface of Shaft. 

Ends perfectly flat, ^ inch diameter. . . Average 2064 lbs. 

Rounded ends, radius | inch • " 2912 ' ' 

Rounded ends, radius J inch " 2573 ' ' 

Cup-shaped and case hardened " 2470 ' ' 

1 10. Shrink and Force Fits. — Relative rotation between 
machine parts is also prevented sometimes by means of shrink 
and force fits. In the former the shaft is made larger than 
the hole in the part to be held upon it, and the metal surround- 
ing the hole is heated, usually to low redness. Because of 
the expansion it may be put on the shaft, and on cooling it 
shrinks and "grips " the shaft. A key is sometimes used in 
addition to this. The coefficient of linear expansion for each 
degree Fahrenheit is 0.0000065 f° r wrought iron and steel and 
0.0000062 for cast iron. Low redness corresponds to about 
6oo° F. and therefore causes an expansion of the bore of about 
0.004 mcn per inch of diameter. Were a hub expanded this 
amount and placed on an incompressible shaft of identical diameter 
so that the fibers immediately surrounding the bore could not 
shrink as the hub cooled, the unit strain in these fibers when 
cool would be the entire .004 inch per inch; and, since unit stress 
equals unit strain multiplied by the coefficient of elasticity, the 
unit stress in the bore fibers would equal .004 X 15,000,000 = 60,000 
lbs. per square inch for cast iron; or .004X30,000,000=120,000 
lbs. per square inch for steel. This indicates that the hub would 
rupture, beginning at the bore fibers, where the unit stress is 
greatest, and extending outward. There is, however, no such 
thing as an incompressible shaft material. Actually the bore 
fibers will not be extended the full amount of the difference be- 
tween the original circumferences at the same temperature of 
the shaft and the hub-bore. The relative compression and ex- 
tension of the two members depends upon several factors such 
as: the value of Poisson's ratio and the coefficient of elasticity 



ME/tNS FOR PREVENTING RELATIVE ROTATION. 179 

(Young's modulus) of the material of each; whether the shaft 
is solid or hollow, and, in the latter case, upon the relation of 
inner to outer diameter; and the relation of diameter of bore of 
hub to its outside diameter. 

It is obvious that in the case of shrink fits the bore expansion 
must be sufficient to allow it to pass freely over the originally 
large shaft to its final position. This in itself precludes the 
possibility of using so large an original difference as .004 inch 
per inch of diameter, even if the certainty of rupture with such 
an allowance did not exist. An allowance of .001 inch per inch 
of diameter, or, as a maximum one of .0012 inch per inch, will 
prove ample and will correspond to as great fiber stresses, with 
ordinary hub proportions, as it is safe to use with either cast 
iron or steel. 

Force fits are made in the same way except that they are 
put together cold, either by driving together with a heavy sledge 
or by forcing together by hydraulic pressure. 

In general, pressure fits are not employed on diameters 
exceeding 10 ins., shrinkage fits being used for large work. 

The alignment should be absolutely accurate in starting. 
To secure this some engineers resort to the use of two diameters 
— each half the length of the fit — differing by but a few thou- 
sandths of an inch, or taper the first half-length an amount equal 
to the allowance. The surfaces should be as smooth as possible 
and well lubricated. Linseed oil is recommended. 

Experience shows that, for same allowances, shrink fits hold 
more firmly than pressure fits.* 

in. Stress in Hub. — In regard to the tension in the hub 
due to shrinkage or forced fits, completely satisfactory data 
are lacking. A close approximation to the probable tension 
in the inner layer of the hub — the one next the shaft — may be 

* For valuable data on fits and fittings, see Am. Mack., Mar. 7, 1907; Trans. 
A. S. M. E., Vols. 24, 34 and 35; Halsey's "Handbook for Machine Designers." 



180 MACHINE DESIGN. 

made by considering the hub as a thick cylinder under internal 
pressure. Using Lame's analysis, Professor Morley * develops 
the following formulas : 

4-' 

' ! *~%-i j_ E\ ri 2 -r2 2 EY ' ' ' W 
w wi £i/ri 2 +r 2 2 ^J 

i>2 = L — , 2) 

m— ii 
mE mi Ei 

P2= , (3) 

2000 

P=^2 ( 4 ) 

/* = unit tensile fiber stress at bore, pounds per square inch; 

p 2 = unit radial pressure on bore surface, pounds per square inch; 

A= total fit allowance; excess of original shaft diameter over 
that of original bore, inches; 

d = original bore diameter, inches; 
/= original length at hub, inches; 

ri = outer radius of hub, inches; 

r 2 = inner radius of hub, inches ; 

£ = Young's modulus, shaft material; 
£1 = Young's modulus, hub material; 

— =Poisson's ratio, shaft material; 
m 

- = Poisson's ratio, hub material; 

* Engineering, Aug. n, 1911. 



MEANS FOR PRESENTING RELATIVE ROTATION. 181 

fi = coefficient of friction between shaft and hub materials; 
^2 = total normal pressure between shaft and hub-bore surface, 

tons. 
P = total pressure required to force shaft into hub, tons. 

Problem. — A steel shaft 5 ins. in diameter is to be forced 
into a cast-iron hub 10 ins. in outer diameter and 8 ins. long. 
Allowing .001 inch per inch excess diameter of shaft over bore 
for the force fit, what will be the unit tensile stress at the bore 
of the hub, and what will be the necessary forcing pressure ? 

A 
J-- 0011 

£=29,000,000 for steel; 
Ei = 14,500,000 for cast iron; 

— =.299 for steel; ^ = 3.35; 

— = .274 for cast iron; mi = 3.65; 

ri = 5"> ^2 = 2. 5"; ^=0.125; 

.001 X 29,000,000 



/.- 



2.35 1 29,000,000X25 — 6.25 29,000,000 \ ' 



^3-35 3- 6 5 i4o 00 > 00 °/ 2 5 + 6 - 2 5 i4,5°°> oco 
= 10,545 lbs. per sq.in.; 

10,545 



.001 — 

14,500,000 



2 -35 + 



3.35 X 29,000,000 3.65 X 14,500,000 
= 6350 lbs. per square inch; 

77X5X8X6350 

P 2 =- =399 tons; 

2000 

P=o.i25X399 = 5o tons, nearly. 



I&2 



MACHINE DESIGN. 



o 1. 

£ 

o 

PQ 

P p. 

S^o. 
° 5 

oft 



S0.4 



,3 

go 



0.2- 



f For U6ual 
in fact in a 


lases with a comparatively small hole in outer ring, 














^j even a considerable hole, the bore of the outer ring expands 
















- (-* 






































n 








































/ 




0a 


ndSc 


lid fc> 
























/ 


/ 


/ 


.,, 


0.2 










0.4 


































/ 






5 








0.6 








































0.7 








































0.8 


















































0.9 






















































1.0 







































0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 

Ratio of Bore Diam. to Outer Diam. of Outer Ring 



1.0 



Fig. 109.4. — Ratio op Expansion of Bore to Force Fit. 
Figures on Curves Denote Ratio of Diameter of Hole in Inner Ring (i.e., Shaft) 
to Diameter of Bore of Outer Ring (i.e., Hub). 



. S-i 



(9 8. 

S3 

MS 
-h o 

w O 



1« 

























14,000 
12,000 




























^ 


d 


^v 










10,000 










___06 




^\ 


f? 


^s 








8,000 










0.7 








<^ 








6,000 












0.8 










% 






4,000 












0.9 
















2,000 


















1.0 

















0.1 



0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 

Ratio of Bore Diam. to Outer Diam. of Outer Ring 

Fig. 109B. — Radial Stress at Bore. Force Fit of o.ooi in. per in. of Bore. 
Stress is directly proportional to Force Fit per Inch of Bore. Figures on 
Curves Denote Ratio of Diameter of Hole in Inner Ring to Diameter of Bore 
of Outer Ring. 



1.0 



MEANS FOR PRESENTING RELATIVE ROTATION. 



183 



112. Hub Stresses by Guest's Law. — Mr. Sanford A. Moss * 
has developed formulae based upon Guest's maximum shear 
law and the use of the same material (steel) in outer and inner 
rings, i.e., hub and shaft. He assumes also that the shaft itself 
may be hollow as well as solid. This theory gives greater values 
to the effective stresses than does the Lame theory. Figs. 109^. 
to 109C show the results in chart form. 









and Solid Shaft 


1 1 1 
















// 




~T" — 


-- — , 


... 


0.2 






















— — 


^__1, 




28,000 








n.:r- 


^-~ 


























1 — i — 1 






0.4 






l^J 


C^ - ' 






















// 


24,000 
a 








0.6 










^^ 


*""~^ 


















/ 


c5 

™ 20,000 

a) 






















>v^ 
















/ 








0.6 




















^ 


^ 


/s 








Ok 












X3 


































II 


v. 






„ lb,U00 






oft- 
























\ 
















ve Str 


































r\ For points 
below this line 
ximum stress 










0.8._ 
























' ffii 












/oti. 


ars in mi 


Eflfec 


















































0.9 
































4,000 





























































































































0.1 



1.0 



0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 

Ratio of Bore Diam. to Outer Diam. of Outer Ring 

Fig. iooC. — Effective Tensile Stress at Bore of Outer Ring. Force 
Fit of o.ooi in. per in. of Bore. 
Stress is Directly Proportional to Force Fit per Inch of Bore. Figures on 
Curve Denote Ratio of Inner Diameter of Inner Ring to Bore Diameter. 



To compute the forcing pressure necessary with these effective 
radial stresses a value of ^=.038, obtained from actual tests, 
is recommended. This low value of /1 indicates that the value 
of the radial stress computed by this theory is too high. The 
same is probably true of the effective tensile stress. 

* Trans. A. S. M. E., Vol. 34. 



1 84 MACHINE DESIGN. 

Fig. 109^4 can be used to determine the approximate unit 
strain for any given materials and proportions of hub and shaft. 
From this strain the corresponding stress can be computed 
directly. In the problem considered in the preceding section 
the ratio of bore diameter to outer diameter of hub = 0.5. The 
shaft being solid, the chart shows a ratio of expansion of bore 
to force fit allowance of .735, found by following up the ordi- 
nate at 0.5 until it intersects the curve for solid shafts. For a 
unit fit allowance of .001 inch, then, the bore fibers in this case 
are strained .000735 mcn P er inch. With a coefficient of elas- 
ticity of 14,500,000, this corresponds to a unit tensile stress of 
10,650 lbs. This checks with the result obtained from Morley's 
formula. Had the shaft been bored out with an internal diameter 
of .5 its outer diameter, the hub remaining the same as before, 
the ratio of bore expansion to unit force fit allowance would have 
been but .59. The unit strain in this case would have been 
.00059 inch per inch, and the corresponding stress, 

14,500,000 X. 00059 = 8550 lbs. per square inch. 



CHAPTER X. 

SLIDING SURFACES. 

113. General Discussion. — So much of the accuracy of 
motion of machines depends on the sliding surfaces that their 
design deserves the most careful attention. The perfection of 
the cross-sectional outline of the cylindrical or conical forms 
produced in the lathe depends on the perfection of form of the 
spindle. But the perfection of the outlines of a section through 
the axis depends on the accuracy of the sliding surfaces. All 
of the surfaces produced by planers, and most of those pro- 
duced by milling-machines, are dependent for accuracy on the 
sliding surfaces in the machine. 




A 

D 



F G 

Fig. no. Fig. hi. 

114. Proportions Dictated by Conditions of Wear. — Suppose 
that the short block A, Fig. no, is the slider of a slider-crank 
chain, and that it slides on a relatively long guide D. The 
direction of rotation of the crank a is as indicated by the arrow. 
B and C are the extreme positions of the slider. The pressure 
between the slider and the guide is greatest at the mid-position, 
A ; and at the extreme positions, B and C, it is only the pressure 
due to the weight of the slider. Also the velocity is a maximum 
when the slider is in its mid-position, and decreases towards the 
ends, becoming zero when the crank a is on its center. The 
work of friction is therefore greatest at the middle, and is very 

185 



1 86 



MACHINE DESIGN. 



small near the ends. Therefore the wear would be the greatest 
at the middle, and the guide would wear concave. If now the 
accuracy of a machine's working depends on the perfection 
of ^4's rectilinear motion, that accuracy will be destroyed as the 
guide D wears. Suppose a gib, EFG, to be attached to A, Fig. 
in, and to engage wiihD, as shown, to prevent vertical loose- 
ness between A and D. If this gib be taken up to compensate 
for wear after it has occurred, it will be loose in the middle 
position when it is tight at the ends, because of the unequal 
wear. Suppose that A and D are made of equal length, as in 

Fig. 112. Then 
when A is in the 
mid-position corre- 
sponding to maxi- 
mum pressure, vel- 

FlG. 112. ._, . 

ocity, and wear, it 
is in contact with D throughout its entire surface, and the wear is 
therefore the same in all parts of the surface. The slider retains 





its accuracy of rectilinear motion regardless of the amount of 
wear; the gib may be set up, and will be equally tight in all 
positions. 

If A and B, Fig. 113, are the extreme positions of a slider, 
D being the guide, a shoulder would be finally worn at C. It 



SLIDING SURFACES. 187 

would be better to cut away the material of the guide, as shown 
by the dotted line, or, better still, to cut a ratchet surface on the 
guide as shown at B. Such a surface aids lubrication by acting 
as an oil reservoir. Slides should always " wipe over " the ends 
of the guide when it is possible. Sometimes it is necessary 
to vary the length of stroke of a slider, and also to change its 
position relatively to the guide. Examples: "Cutter-bars" 
of slotting- and shaping-machines. In some of these positions 
there will be a tendency, therefore, to wear shoulders in the 
guide and also in the cutter-bar itself. This difficulty is over- 
come if the slide and guide are made of equal length, and the 
design is such that when it is necessary to change the position 
of the cutter-bar that is attached to the slide, the position of 
the guide may be also changed so that the relative position of 
slide and guide remains the same. The slider surface will then 
just completely cover the surface of the guide in the mid-position, 
and the slider will wipe over each end of the guide whatever 
the length of the stroke* 

In many cases it is impossible to make the slider and guide 
of equal length. Thus a lathe-carriage cannot be as long as the 
bed, a planer-table cannot be as long as the planer-bed, nor a 
planer-saddle as long as the cross-head. When these condi- 
tions exist especial care should be given to the following: 

I. The bearing surface should be made so large in pro- 
portion to the pressure to be sustained that the maintenance 
of lubrication shall be insured under all conditions. 

II. The parts which carry the wearing surfaces should be 
made so rigid that there shall be no possibility of the localiza- 
tion of pressure from yielding. 

115. Form of Guides. — As to form, guides may be divided 
into two classes: angular guides and flat guides. Fig. 114, a, 
shows an angular guide, the pressure being applied as shown. 

* See further, Things That Are Usually Wrong, by Prof. J. E. Sweet. 




1 88 MACHINE DESIGN. 

The advantage of this form is, that as the rubbing surfaces 
wear, the slide follows down and takes up both the vertical 
and lateral wear. The objection to this form is that the pres- 
sure is not applied at right angles to 
the wearing surfaces, as it is in the 
flat guide shown in b. But in b, a gib 
must be provided to take up the lateral 
wear. The gib is either a wedge or 
a strip with parallel sides backed up by screws and offers 
danger of localized pressure or binding. Guides of these forms 
are used for planer-tables. The weight of the table itself holds 
the surfaces in contact, and if the table is light the tendency 
of a heavy side cut would be to force the table up one of the 
angular surfaces away from the other. If the table is very heavy, 
however, there is little danger of this, and hence the angular 
guides of large planers are much flatter than those of smaller 
ones. In some cases one of the guides of a planer-table is 
angular and the other is flat. The side bearings of the flat guide 
may then be omitted, as the lateral wear is taken up by the 
angular guide. This arrangement is undoubtedly good if both 
guides wear down equally fast. 

As regards lathe beds, while English practice favors the flat 
way, the form of guide used in America is almost exclusively 
the inverted V. The angle for small lathes is oo° and increases 
slightly in obtuseness with the larger sizes. The main advantage 
of the inverted V form is that of automatic, certain adjustment, 
the adjustment with flat ways depending upon the use of gibs. 
On the other hand, the flat ways offer a more distributed sup- 
port to the carriage, which is therefore less apt to spring under 
the stress of a heavy cut than if supported solely at the outer Vs. 
Fig. 114A shows a modern compromise as adopted on a small lathe. 
Fig. 115 shows several forms of sliding surfaces such as are 
used for lathes, shapers, milling machines, etc. 



SLIDING SURFACES. 



189 




S 



L UU[j_ 



% x Ha- 

parallel gib 
y> set screw-- 
and lock nut 



Fig. 114^4 



AL 



7 



ill! 



STTT-J 




|^ 








Fig. 



190 MACHINE DESIGN. 

At A is shown a form employing a taper gib, usually made 
of bronze, adjusted by means of a stud and lock nuts, d. The 
angle is generally 6o°, although it is sometimes made as acute 
as 45 . The form shown at B also employs a taper gib, but it 
is set up by means of a headless screw engaged in a thread cut 
in the slide. The threads are cut out of the corresponding 
portion of the gib, and the end of the screw, pressing against 
the gib at the bottom of this recess, forces the gib in as the screw 
is turned. The adjustments of C, D, and E are obvious. These 
heavier gibs are usually made of cast iron; they are not tapered 
lengthwise. C is the most accessible, but least stable. F em- 
ploys a cast-iron gib, tapered and adjusted by means of a screw 
which has a broad collar or disk. G and H employ parallel 
thin gibs of bronze or steel, set up by means of adjusting screws. 
This form is not so satisfactory as the wedge gib, as the bearing is 
chiefly under the points of the screws, the gib being thin and yield- 
ing, whereas in the wedge there is complete contact between the 
metallic surfaces. / shows an arrangement, employing no angular 
surfaces, such as is used on some shaping machines to constrain the 
ram. K shows an arrangement employed on planer cross-rails. 

The sliding surfaces thus far considered have to be designed 
so that there will be no lost motion while they are moving 
because they are required to move while the machine is in 
operation. The gibs have to be carefully designed and accu- 
rately set so that the moving part shall be just " tight and 
loose "; i.e., so that it shall be free to move, without lost motion 
to interfere with the accurate action of the machine. There is, 
however, another class of sliding parts, like the sliding-head 
of a drill -press, or the tailstock of a lathe, that are never required 
to move while the machine is in operation. It is only required 
that they shall be capable of being fastened accurately in a 
required position, their movement being simply to readjust them 
to other conditions of work while the machine is at rest. No 



SLIDING SURFACES. IQI 

gib is necessary and no accuracy of motion is required. It 
is simply necessary to insure that their position is accurate 
when they are clamped for the special work to be done. 

116. Lubrication. — The question of strength rarely enters 
into the determination of the dimensions of sliding surfaces; 
these are determined rather by considerations of minimizing 
wear and maintaining lubrication. As long as a film of oil 
separates the surfaces, wear is reduced to a minimum. The 
allowable pressure between the surfaces without destruction 
of the film of lubricant varies with several conditions. To 
make this clear, suppose a drop of oil to be put into the middle 
of an accurately finished surface plate (i.e., as close an approxi- 
mation to a plane surface as can be produced) ; suppose another 
exactly similar plate to be placed upon it for an instant; the 
oil-drop will be spread out because of the force due to the weight 
of the upper plate. Had the plate been heavier it would have 
been spread out more. If the plate were allowed to remain 
a longer time, the oil would be still further spread out, and if 
its weight and the time were sufficient, the oil would finally be 
squeezed entirely out from between the plates, and the metal 
surfaces would come into contact. The squeezing out of the oil 
is, therefore, a function of the time as well as of pressure. 

If the surfaces under pressure move over each other, the 
removal of the oil is facilitated. The greater the velocity of 
movement the more rapidly will the oil be removed, and there- 
fore the squeezing out of the oil is also a function of the velocity 
of tlie rubbing surfaces. 

117. Allowable Bearing Pressure. — Flat surfaces in machines 
are particularly difficult to make perfectly true in the first place, 
and to keep true in the course of operation of the machines. 
If they are distorted ever so slightly the pressure between the sur- 
faces becomes concentrated at one small area, and the actual pres- 
sure per square inch is vastly in excess of the nominal pressure. 



192 MACHINE DESIGN. 

In consequence of this and the differences in original truth 
and finish of the surfaces, there is no matter in machine design 
in which practice varies more than in the nominal pressure 
allowed per square inch of bearing area of flat sliding surfaces. 

Unwin * gives the following : 

Table XIV. 

Cast iron on babbitt metal 200 to 300 lbs. per square inch 

Cast iron on cast iron, slow 60 to 100 lbs. per square inch 

Cast iron on cast iron, fast 40 to 60 lbs. per square inch 

Professor Barr f found American practice to vary as follows : 

Cross-head shoes of high-speed engines: 

Minimum pressure per square inch 10 . 5 lbs. 

Maximum pressure per square inch 38 " 

Mean pressure per square inch 27 ' ' 

Cross-head shoes of low-speed engines: 

Minimum pressure per square inch 29 lbs. 

Maximum pressure per square inch 58 ' ' 

Mean pressure per square inch 40 " 

In all cases the mean sliding velocity was probably in the 
neighborhood of 600 feet per minute with a maximum velocity 
at the middle of the stroke of about 950 feet per minute. In 
" low-speed " engines the maximum velocity is only reached 
about one third as many times per minute as in " high-speed " 
engines, although they may have the same mean velocity, and 
it is therefore proper to allow a higher unit value of pressure 
for the former than for the latter. For well-made surfaces the 
maximum values given by Professor Barr may be safely used. 

For lower mean speeds than 600 feet per minute they may 
be increased, and for higher speeds decreased, according to some 
law such as ^7=36,000, 

in which formula p = pressure per square inch, and V= velocity 
of rubbing in feet per minute. 

* "Machine Design," 1909, Vol. I, p. 252. 
t Trans. A. S. M. E., Vol. XVIII, p. 753. 



SLIDING SURFACES 



193 



118. Maintenance of Lubrication.— Regarding the materials 
to be used, brass, bronze, or babbitt metal will run well with 
iron or steel. Cast iron on cast iron is frequently used, par- 
ticularly in machine tool construction. 

To maintain lubrication a constant flow of oil from a cup is 
desirable. The moving surface should, if possible, have chan- 
nels cut in its face to conduct the oil from the central oil-hole 
to all parts of the surface, as shown in Fig. 116. The layout 
of the oil-grooves should be such as will take advantage of the 




Fig. 116. 



I I 




Fig. 117. 



Fig. 



motion of the part to draw the oil along them and distribute 
it over the entire bearing area. The oil should be forced in where 
the pressures are heavy. 

Oil-pads may be used as shown in Fig. 117. The shaded 
areas represent porous pads whose lower sui faces just touch the 
surface to be lubricated, and which are kept soaked with oil. 

For oiling the ways of planer-tables it is customary to use 
rollers placed in oil- filled pockets in the guides. The top of 
the roller is held against the surface of the way by means of 
springs. (See Fig. 118.) 

It is extremely difficult to maintain continuous film lubri- 
cation with reciprocating surfaces except with forced lubrication. 
In some cases it may be possible to employ the principle of the 
Kingsbury thrust bearing. (See p. 246.) 



CHAPTER XI. 



AXLES, SHAFTS, AND SPINDLES. 

119. By Axles, Shafts, or Spindles we denote those rotating 
or oscillating members of machines whose motion is constrained 
by turning pairs. Axle is the name given to such a member 
when it is subjected to a load which produces a bending moment, 
and the only torsional stress is that due to friction. 

When rotating members are subjected chiefly to torsional 
stress,' or combined torsion and bending, they are called shafts 
or spindles. The former term is used where the part has as 
its function the transmission of energy of rotation from one 
point to another. Examples are line-shafts and crank-shafts. 

The term spindle, on the other hand, is restricted to those 
rotating members which are directly connected with the tool or 
work and give it an accurate rotative motion. They generally 
form the main axis of the machine. Examples are the lathe- and 
drill-spindles. 

120. Axle Design. — The question of axle design will be taken 
up first, and the torsional moment due to friction will be neglected. 



t- 




p 


Q 




S 


-t- 


• 


> 









Fig. 119. 

A typical case is shown in Fig. 119. Here the two ends are 
purposely not symmetrical. Given the loads P and Q, solution 
is first made for the reactions R and S by the ordinary methods 
of mechanics, grauhical or analytical. 

194 



AXLES, SHAFTS, AND SPINDLES. 



195 



The graphical method is best, since it gives the moments at 
all sections. Lay off the line M-N, Fig. 120, whose length 
equals /' + / + /", the distance between the points of applications 
of R and S. Denote the points of application of R, P, Q, and 
S by a, b, c, and d, respectively. At b erect a perpendicular and 




Fig. 120. 

lay off on it a vector representing the 
value of P in pounds. At c erect a 

perpendicular and lay off on it a vector representing Q on the 
same scale. At d drop a perpendicular and lay off de equal 
to vector Q, and ef equal to vector P. Select any point O 
as pole, and draw Od, Oe, and Of. Denote by g the point 
where Od intersects a perpendicular dropped from c, and draw 
from g a parallel to Oe until it intersects a perpendicular dropped 
from b at h. From h draw a parallel to Of until it intersects a 
perpendicular dropped from a at /. Draw jd, and parallel to jd 
draw a line through O. This line cuts the perpendicular dropped 
from d at the point k. Then vector fk = R, and kd = S, on 
the same scale as was originally used for P and Q. Values 
of R and S in pounds are therefore determined. The shaded 
area dghjd is the moment diagram. The vertical ordinates 
included between its bounding lines are proportional to the 
moments at the corresponding points. 

The scale of the moment diagram can be readily determined 
by solving for the actual moment for one section. Select the sec- 
tion at b. The moment Mb is represented by mh and has a 
value =Rl', R being expressed in pounds and /' in inches; the 



196 MACHINE DESIGN. 

value of the moment in inch-pounds can be determined at any 
point, since the scale used is mh inches equals Rl' inch-pounds. 
For a circular section we have the elastic moment 



M 



III ftKf 3 



c 4 

Mb is the bending moment in inch-pounds; 
ft is the unit stress in outer fiber in pounds per square inch; 
/ is the plane moment of inertia of the section in biquad- 
ratic inches; 
c is the distance from neutral axis to outermost fiber in 
inches. 

Equating this to the various selected values of Mb and solv- 
ing for r gives the radius of the axle at any point. 

In designing axles, great care must be taken that all forces 
acting are being considered, and that the maximum value of 
each is selected.* 

Thus it has been found that the force due to vertical oscilla- 
tion caused by jar in running is about 40 per cent of the static 
load for car-axles. The axles would therefore have to be de- 
signed for a load 1.4 times the static load. In addition to this 
there is in car-axles a bending moment due to curves, switches, 
and wind-pressures. This may amount as a maximum to the 
equivalent of a horizontal force H, equal to 40 per cent of the 
static load, applied at a height of 6 feet from the rail. 

When such a careful analysis of the forces has been made, 
} t may be taken for good material, equal to one fourth of the 
ultimate strength, this being a safe value for cases like this, 
where the fibers are subjected to alternate tension and com- 
pression as determined by Wohler and others. 

Had the static load alone been considered in the calculations, 

* See further Proc. Master Car-Builders' Assn., 1896; Report of Committee 
on Axles, etc. Also Strength- of Railway-Car Axles, Trans. A. S. M. E., 1895. 
Reuleaux, "The Constructor," trans, by H. H. Supplee, Philadelphia, 1893; 
Railway Machinery, Mar. 1907. 



AXLES, SHAFTS, AND SPINDLES. 197 

} t should not have been taken greater than one tenth of the 
ultimate strength. 

121. Shafting Subject to Simple Torsion. — If a short shaft 
is subjected to simple torsion, its diameter may be deter- 
mined very readily by the simple formula for torsional 
moment, 

M,=Pa= f —. 

c 

Here M t = Pa is the torsional moment, P being the force tending 
to twist the piece in pounds and a being the lever-arm of P 
about the axis of the piece in inches. 

/ is the polar moment of inertia of the cross-section of the 

member in biquadratic inches; 
c is the distance from the neutral axis to the outermost fiber 

in inches; 
/« is the allowable unit stress in pounds per square inch. 
For a solid circular section 

/ nr 3 
c 2 

and M t = , 

2 

which can be solved readily for r, the radius of the shaft. 
For a hollow circular (i.e., ring-shaped) section, 



and M t = 



C 2Ti 



2Y X 



Here r\ is the radius of the outside of the shaft and r 2 is the 
radius of the bore, both in inches. 

The way to solve this is to take r 2 as a decimal part of f\. 
Thus, let r 2 = bri. It then becomes an easy matter to solve for r lt 



198 



MACHINE DESIGN. 



122. Shafting Subject to Combined Torsion and Bending. — 

In most cases shafts are subjected to combined torsion and 
bending. Consider the crank-shaft shown in Fig. 121 in side 
and end view. 



fl c 




B 


X 










y 




p 




P 






— 1 


M 




Fig. 121. 



B is the center of the bearing, C is the center of the crank- 
pin. At B we have the shaft subjected to a bending moment, 
Mb=Pl, and also to a twisting moment, M t =Pa. 

Let M e b represent the bending moment which would produce 
the same stress in the outer fiber as Mb and M t combined. It 
will be called the equivalent bending moment. Then it has 
been found * that 



M eb = o. 3 sM b + o.6sVM b 2 + M t 2 (1) 



Also, 



Ma,= 



fl 



(2) 



fizr^ 



For a circular section (2) becomes M e *= ; and substitution 



in (1) gives 



fnr* 



= o. 3 5M b +o.6sVM b 2 + M?,. . . (3) 



* See Bach, "Elasticitat und Festigkeit.' For simplicity's sake, the coefficient 
«o has been dropped, as it modifies the result very slightly for wrought iron and 
steel. This is based upon St. Venant's maximum strain theory. Guest (Phil. 
Mag., July, 1900, p. 132) advances the formula Meb—^Mb 2 -\-Mt 2 as based upon 
the maximum shear theory and experiments on combined stresses, but states 
(p. 78) that experimental results cannot be held to disprove the maximum strain, 
theory as they do (p. 77) the maximum stress, or Rankine, theory. 



AXLES, SHAFTS, AND SPINDLES. 199 

which can readily be solved for r, f being given a value equal 
to the maximum allowable unit tensile stress for the material 
and conditions.* 

For a hollow circular section 

A(ri 4 ~ r ^ = . 3S M b +o.65VM b 2 + M?. ... (4) 

To solve this express r 2 as a decimal part of Y\. Substitute and 
solve for n. 

If there are several forces acting, as there are apt to be, the 
method is as follows: First, find Mb due to all the bending 
forces combined. Second, find M t due to all the twisting forces 
combined. Third, use these values of Mb and M t in equations 
(1), (3), and (4). Among the forces acting we must not fail 
to include the weight of the shaft and attached parts. 

123. Comparison of Solid and Hollow Shafts. — It is evident 
from (3) and (4) that the dimensions of a solid shaft and a 
hollow shaft of equal strength will have the relationship 



If r2 = o.6ri, we have 



3/ ri 4_ r2 4 

r=o.955ri. 



* Merriman, "Mechanics," nth ed., p. 267, develops formulae on the maxi- 
mum shear theory which reduce to 



— = M b + VM t 2 +Mb 2 , 
2 

and UI^ = y/Mt*+Mb 2 , 

2 

in which /', is the unit resultant shear on diagonal surface due to combination of 
unit flexural tension and unit torsional shear, and /' is the unit resultant tensile 
stress normal to the diagonal surface. According to Guest's results, f' s must have 
a value less than one-half of the unit tensile strength at "yield point" for ductile 
materials. 



200 MACHINE DESIGN. 

Hence a hollow shaft whose internal bore is 0.6 of its ex- 
ternal diameter, in order to have the same strength as a solid 
shaft must have its external diameter 1.047 times the diameter 
of the solid shaft. The weight of the hollow shaft will be 70 
per cent of that of the solid shaft. It is obvious that a con- 
siderable saving in weight may be effected without appreciable 
increase in size if the hollow section is adopted. By using nickel 
steel -in connection with the hollow section we can get com- 
bined maximum strength and lightness.* 

124. Angular Distortion. — The angle by which a shaft sub- 
jected to torsion is twisted is often an important matter. Let 
this angle be represented by ■&. Then 

M,h8o 

»=-JaT- ■ ■ ■ (s) 

# = angle of torsion in degrees; 
M t = twisting moment in inch-pounds, 
/= length of shaft in inches; 

/ = polar moment of inertia in biquadratic inches; 
modulus of elasticity 



G = modulus of torsion : 



2.6 



125. Combined Pull or Thrust and Torsion. — When a shaft 
is subjected to combined tension and torsion, or compression 
and torsion, the following formulae have been developed by 
Prof. Merriman : f 

7id?f'= 2 Pd+ V(i6M t ) 2 + (2Pd) 2 , 



7cd?f' 8 = V(i6M t ) 2 + ( 2 Pd) 2 . 

P = total axial load, tension or compression, pounds; 
/' = unit normal resultant tensile or compressive stress, 
pounds per square inch; 

♦See "Nickel Steel," a paper by D. H. Browne in Vol. 29 of the Trans. 
A. I. M. E. 

f Mechanics, Wiley & Sons, 1914, p. 268. 



AXLES, SHAFTS, AND SPINDLES. 20 1 

// = unit diagonal resultant shear stress, pounds per square inch; 

M t = twisting moment, inch-pounds; 
d = diameter of shaft, inches. 

In compression cases these formulae are applicable strictly 
only to vertical shafts with supports sufficiently close together to 
make it legitimate to consider them as short columns. 

The value of d is found by substituting trial values in the 
equations until one is determined which satisfies the identity. 

When the distance between bearings is such that the shaft, 
subjected to combined thrust and torsion, must be considered 
as a long column, the following formula has been developed 
by Prof. A. G. Greenhill:* 

7Z 2 P M? 

+ 



I 2 EI 4&I 2 ' 

/=the length of shaft between bearings in inches; 
P = the end thrust in pounds; 
E = modulus of elasticity; 

7= plane moment of inertia of the section in biquadratic inches; 
M t = twisting moment in inch-pounds. 

This formula, also, is applicable strictly only to vertical shafts, 
as it ignores the important item of bending due to the weight 
of the shaft and attached parts. 

126. Line-shafts. — Line-shafts are long shafts used to trans- 
mit power. They are made of lengths coupled together and 
supported by bearings at suitable intervals. Pulleys or gears 
are keyed to them, and should always be placed as close to the 
supporting bearings as possible. 

If line-shafting is subjected to simple torsion the equation 

M t = — -jr- may be used to determine its diameter. It is frequently 
* See Proc. of Inst, of M. E., 1883, p. 182. 



202 MACHINE DESIGN. 

convenient in such a case to express M t , the twisting moment 
in inch-pounds, in terms of the horse-power to be transmitted, 
H.P., and the number of revolutions per minute of the shaft, N. 

Then, H.P.X 33,000X12 = inch-pounds of work per minute 
= 2nNM t} 

33,000X12 H.P. H.P. 

■'■ Mt= — ST" -AT =63 '° 24 -iv-' 

and , 3= i6X63P2 4 H_P. 

nf s N 

If f s = gooo lbs. per square inch for mild steel, repetitive 
stress between o and maximum, one direction only, 



s/H.P. 



N 

For ordinary line-shaft, where there is an average amount 
of bending as well as torsion, the common rule of practice is 



3 / 



H.P. 



N 
For prime mover shafts and jack-shafts the rules of practice 

3 (h.pT 3/n."py 

for d range from 4.64^—-— to 5-^——-, when the bearings are 

close to main sheaves or pulleys. 

In other cases it is better to compute the diameter by equa- 
tion (3) for combined torsion and flexure, choosing the value 
of / according to the material to be used and the conditions of 
stress application and variation, i.e., in regard to maximum 
and minimum repetitive stresses, reversals, and gradual or sudden 
application of load. 

If line-shafts are designed wholly for strength, owing to 
their length there is apt to be an excessive angular distortion #. 
It is therefore desirable to design them for stiffness and check 
for strength afterwards. 



AXLES, SHAFTS, AND SPINDLES. 203 

t? should not exceed 0.07 5 per foot of shaft. 

Combining this rule with the formula (5) for angular distortion, 

M,/i8o 
for a round wrought-iron or steel solid shaft this gives 

when d= diameter of shaft in inches; 

H.P.= horse-power to be transmitted; 
N = revolutions per minute of shaft. 
Having determined the diameter which will give sufficient 
stiffness against torsion the allowable distance between sup- 
porting bearings must be calculated. The rule of practice is 
to limit the deflection to 1/100 of an inch to a foot of length. 

Consider first a bare shaft supported at both ends. There 
are three cases : 

1st. Both ends of the shaft are free to take any direction. 
2d. One end is free and one fixed. 
3d. Both ends are fixed. 
In each case / = length of span in inches; 
w = weight of shaft per inch; 
y = maximum deflection; 

24/y 

— = average deflection per foot of length; 

= 1/100 inch; 

/ 

/. y= . 

2400 

Each case is that of a uniformly loaded beam with a loa,d= wl. 

wl 4 
For case I, the deflection y = — — . 

77EJ 

wl 4 
For case II, the deflection y= 



iS S EI 



204 MACHINE DESIGN. 

For case III, the deflection y = . 

384^7 

Since y= , and for round shafting 1=—- and w= 

2400 04 

7rd 2 
.28 — , while £=29,000,000, it follows that 

4 

(Case I) 1= 6o^J 2 ; 

(Case II) 1= &o«VJ 2 ' f 

(Case III) /= 103^. 

When there are loads due to belt pull, etc., the deflection 
must be determined in each case. For ordinary purposes, 
with the average number of pulleys and amount of belt pull, 
it is safe to take for loaded shafts T % of the value of / deter- 
mined for bare shafting for the same conditions. Case II 
corresponds most closely to ordinary conditions. 

127. Critical Speed of Shafts. — It is a practical impossibility 
to have the center of mass of a rotating shaft lie in the mechan- 
ical axis of rotation. In the case of a horizontal or inclined 
shaft it is obvious that there will be a deflection, due to its own 
weight between supports even if there is no additional load, 
which will cause- the center of mass to lie below the true axis 
of rotation. As the shaft begins to rotate, because its center 
of mass lies off the axis of rotation, it will be subjected to a 
<v centrifugal force " tending to make it take a bowed form in 
rotating. At first this tendency is not sufficient to overcome the 
tendency to remain bent downward, but as the speed increases 
the " centrifugal force " increases until a state of equilibrium is 
reached, at what is called the critical speed, when the centrifugal 
force will be just sufficient to counteract the force tending to 
deflect the shaft downward and the shaft will rotate in a bowed 
form. It is then said to whirl. 

In the case of vertical shafts the rotating body would have 



AXLES, SHAFTS, AND SPINDLES. 205 

to be in perfect balance, which is only theoretically possible, 
otherwise — and this is always the case — the center of mass lies 
off the mechanical axis. As the speed increases there will be the 
same conflict between centrifugal force, tending to bow the shaft 
more and more, and the elastic resistance of the material. This 
action continues until a state of equilibrium is reached at which 
the force of the shaft deflection {i.e., the force which the shaft 
is capable of exerting at its mass center in its effort to return to 
its original straight form) is equal and opposite to the " centrif- 
ugal force " of the mass. This will be at the same critical speed 
as in the horizontal case. 

Theoretically the bow of the shaft becomes indefinitely great 
at the critical speed. Practically this is not so. The speed may 
be increased very greatly beyond this. But at the critical speed 
there will occur the maximum vibration of the revolving mass 
and consequently of its supports. The vibrations are smaller 
for both higher and lower speeds. The determination of the 
critical speed is, therefore, a matter of prime importance in the 
design of all high rotative speed machinery. 

Above the critical speed the center of mass revolves i aside 
the bow of the shaft; and the tendency of the rotating mass is 
to rotate about an axis through its own center of gravity and 
not about the mechanical axis. Designers provide for this either 
by the use of a flexible shaft (made just strong enough to with- 
stand the deflection stresses in passing through the critical speed), 
or by the introduction of special bearings permitting this ac- 
commodation and constructed to dampen the vibration effect. 

The critical speed here spoken of is the first or lowest critical 
speed, N\. There is a series of secondary critical speeds, N, 
of higher value with diminishing amplitudes of vibration. 

The mathematical treatment is too long to be introduced 
here but can be found in its simplest form in a paper by Mr. 
S. H. Weaver in the Jour, of the A. S. M. E., June, 1910, from 



206 



MACHINE DESIGN. 



Critical Speed Formulae. 

Weights in Pounds, Dimensions in Inches, Vertical Shafts Considered 

Horizontal. 
N, Ni, N 2 = critical speeds in r. p. m. 

Ai, A 2 = static deflections at W\ and W2 (shaft horizontal). 
d = diameter of shaft (inches). £=29,000,000. 



Single concentrated Load. 
General Formula 



E 



-2 



-b- 



1 



VAi 



N 



*-i87-7-\£ 



iVi = 387,000^ 
A Wia*b 2 



\K 



3^ 



W, 



T± 



fctltrtj 



iV x = 1,550 



iVi= 187.7 \/ s 



5 °°^ 2 \fe 



Ar 



W 3 
: 4 8£/ 



w, 



J, 



*-.-*-*=* 



2^=387,000^^ 



^1 = 187.7 
3 EIP 



A,= 



iVi = 3,100,850^ 






-r-3E^f 



JVi=i87 



' 7 \£ 



\Wii 



IQ2 EI 



w. 



^ 



■ r HwJ^ 



iV 






A WW, , , ,, 









i 2 - 



3 



<2 2 

iVi = 2,33 7,00c— 



tt- 187.7 <fe 



7 w 3 



Vwy 



Ai 



A,= 



768 £/ 



AXLES, SHAFTS, AND SPINDLES. 
Critical Speed Formulae. — Continued. 



Two Concentrated Loads 
General Formulae 



E^ 



.„{. 



fcr_.,_ ^ 



j T , 



-a->U 



t+i-i^it^ 



^ = 387,000^ 



Wi/ 3 



^1 = 187.7 

Al= 3^ 



xS 



2Vi and iV 2 



//*2Ti , M l/JSTi X 2 \ 2 , 4^3 2 



6£/ 



(/— ai— a 2 ) 2 [/(3/— 2^—202) -(a 2 —ai) 2 } 



K^C-^Kl-azY 



K 2 = C—Ml-a x y 



<J2 



K _ J(l 2 —(h 2 —ai 2 ) — aia2(ai — <h) 

dl02 

A T i and N 2 = Substitute in Equation (16) 



if 2 / 1 

iV 2 =548, 4 oo^ pFi(3/ _ 4ai) 

Ai== i£r (3 ^- 4fli) 



4/i/2(fi+W-/i(/i+a2)»-k(/ I +ai) 2 
#1 = (^) C[4/ 2 (/i+/ 2 )-(/ 2 +a2) 2 ] 

K 2 = (J£j 'ciMk+^-di+a^] 



hfa 



■C(h+a 1 )(l 2 +a 1 ) 



ai&i02&2 

Ni and A 7 2 = Substitute in Equation (16) 



3EII 



2a 2 b 2 W 2 -{l+a) 2 \ 
K: = K, = C[8l 2 -(l+a) 2 } 
/C 3 = C(/+a) 2 
A T i and N 2 = Substitute in Equation (16) 



207 



208 



MACHINE DESIGN. 
Critical Speed Formula. — Continued. 



K3p?^ 



w, 



U It T"~h 



W, 



J 



w„ 



U " 



2 -r 2 
< II H*~- ly-H 



Distributed Loads 



Total Load =W 



F 



#1=1,547,000^-^— 

^1=124.3 Vt 

^2=2,337,000^1^ 
^ = 187.7^- 
7 l^ 3 



r= 12E7 

aVU^i+y-tfi 2 -^ 2 ) 2 ] 

K 2 = CaWh 

Kz=-a!Mli 2 -a 2 ) 
2 

Ni and iV 2 = Substitute in Equation (16) 
. Wia 2 b 2 ah n - ,x 

A ^^EIh- W *6ETh {1 ' ' 0) 
A2 = g! ( , + «_^ (/l2 _ oJ) 



3 £/ 



Ki = i6Ch 2 (li+h) 

K z =3Ch% 

Ni and iV 2 = Substitute in Equation (16) 

WW W^ 2 



A 2 = 



48£7 16 EI 
WMh+k) Wrhh* 



3EI 



16EI 



A = Maximum Static Deflection 



#! = 2,232,510^-^3 
iVi=2ii.4-\/^ 



Ni =4,760,000— (Shaft alone) 

N = [i, 4, 9, 16, etc.]Ni 
5 TFZ 3 



A = 



384 EI 



AXLES, SHAFTS, AND SPINDLES. 
Critical Speed Formulae. — Continued. 



209 





#l=4,979,250d 2 -y-j-3 


55 


1 


Total Load = W 
iliillliiiilll] 


i 


#i=24sVf 

Ni = 10,616,740— (Shaft alone) 


56 


« I ^ 


! 


57 


' 


N = [i, 2.78, 5.45, 9, etc-JiVx 


58 




WP 

3&4EI 


59 




#1=795,196^^-3 


60 


Total Load =W 


Ni = 167.6 -\J^ 

Ni = 1,695,514 jy (Shaft alone) 


61 


■Uuiuunmn 


62 




# = [1,6.34, 17.6, 43.6, etc.]#L 


63 




a WP 
A = 8ET 


64 




^ = 3,482,715^/^3 


65 


1 


Total Load =W 

iinmumn 


-A 
J 


#1 = 209.7 ^ 

Ni = 7,021,600 — (Shaft alone) 


66 


1 


., , 


67 




N=[i, 3.24, 6.8, 1 1.6, etc.] #1 


68 




WP 
185E/ 


69 



which the appended convenient tables are abstracted. In addi- 
tion to these tables the following formula * for a multiple loaded 
shaft will be found useful. 



N x 



NaNJVe 



y/N b *N* + N 2 N 2 + N 2 N t 



Ni = first critical speed of the system; 
N a = first critical speed of unloaded shaft ; 
N b , N c = first critical speed of each separate load. 



* E. A. Lof in Machinery, Feb., 1909. 



CHAPTER XII. 

JOURNALS, BEARINGS, AND LUBRICATION.* 

128. General Discussion of Journals and Bearings. — Jour- 
nals and the bearings or boxes with which they engage are 
the elements used to constrain motion of rotation or vibration 
about axes in machines. Journals are usually cylindrical, but 
may be conical, or, in rare cases, spherical. The design of 
journals, as far as size is concerned, is dictated by one or 
more of the four following considerations. 

(1) To provide for safety against rupture or excessive yield- 
ing under the applied forces. 

(2) To provide for maintenance of form. 

(3) To provide for maintenance of lubrication. 

(4) To provide against overheating.. 

To illustrate (1), let Fig. 122 represent a pulley on the 
end of an overhanging shaft driven by a belt, ABC. Rota- 
tion is as indicated by the arrow, and the belt tensions are T 1 
and T 2 . The journal, /, engages with a box or bearing, D. 
The following stresses are induced in the journal: Torsion, 
measured by the torsional moment {T 1 -T 2 )r. Flexure, 
measured by the bending moment {Ti + T 2 )a. This assumes 
a rigid shaft or a self-adjusting box. Shear, resulting from 
the force T 1 + T 2 . This journal must therefore be so designed 
that rupture or undue yielding shall not result from these 
stresses. The method of doing this is outlined in later sections 
of this chapter, 

To illustrate (2), consider the spindle journals of a grind- 
ing-lathe. The forces applied are very small, but the form 

* See further, Vol. 27 Trans. A. S. M. E., pp. 420-505. 



JOURNALS, BEARINGS, AND LUBRICATION. 



211 



of the journals must be maintained to insure accuracy in the 
product of the machine. A relatively large wearing surface 




Fig. 122. 

is therefore necessary, and careful provision must be made to 
exclude dust and grit. Journals whose maintenance of form is 
of chief importance must be designed from precedent, or accord- 
ing to the judgment of the designer. No theory can lead to 
correct proportions. In fact these proportions are eventually 
determined by the process of machine evolution. 

129. Journal Friction and Lubrication. — Consideration (3). 
When two solid surfaces have relative motion under pressure 
there will be friction between them. The ratio of this frictional 
resistance, F, to the normal pressure, P, is termed the coefficient 
of friction, fi. 

Cylindrical journals having a diameter = d inches and length 
= / inches, have a projected area = d/ square inches. In what 
follows the area of a journal means its projected area. Thus 

P 



the pressure per square inch of journal = p 

Cbl/ 

there are three kinds of friction to be considered. 



In machinery 



212 MACHINE DESIGN. 

i. The surfaces are dry. — In this case the ordinary laws 
of solid (dry) friction apply. With very smooth and clean sur- 
faces cohesion may take place and cold welding result. However, 
the friction between most so-called unlubricated surfaces is not 
a case of true friction between pure metals, but between sur- 
faces contaminated by atmospheric agencies, by grease, etc., 
derived from handling or from the material with which the 
surfaces were wiped, or by chemically formed films such as 
oxides, sulphides, etc. In other words the surfaces are partially 
lubricated. Since the extent of contamination varies it is clear 
why different experimenters have obtained conflicting results 
(even when using materials of like character as regards surface 
conditions, hardness, etc.) and have deduced differing laws. 

For approximately clean, dry, metallic surfaces: 

(a) The frictional resistance is approximately proportional to 

the load. 

(b) The frictional resistance is slightly greater for large areas 

and small pressures than for small areas and large 
pressures. 

(c) The frictional resistance, with the possible exception of 

very low speeds, decreases as the velocity increases. 

With regard to machinery these laws apply to some friction 
clutches and brakes and also to those bearings which are not 
lubricated but have the journal running in bushings of graphite 
or chemically treated wood. Such bearings have been found to 
give very satisfactory service where the unit pressures are moderate 
and where it is either difficult to apply oil or undesirable to use 
it, as in textile manufactures. These laws also apply in the 
starting of heavy machinery where the bearing surfaces come into 
metallic contact owing to their having lain at rest a sufficient 
time under pressure adequate to force out the lubricant, even 
though the natural action of running tends to reintroduce it be- 



JOURNALS, BEARINGS, AND LUBRICATION. 213 

tween them later. In the case of metallic contacts the static 
coefficient of friction of slightly contaminated surfaces is the 
highest value their coefficient of friction can have. For the 
ordinary materials used for journals and boxes static n ranges 
in value from .14 to .22. 

In the case of a soft surface on a hard one, such as a leather 
belt on a cast-iron pulley, these laws do not hold strictly. In 
such cases it has been found * that the coefficient of friction 
starting with a static value of say .12 increases with velocity of 
slip until it reaches a maximum considerably greater than unity 
at a speed of slip of 600-700 feet per minute. 

2. The surfaces are partially lubricated. — Here the investigator 
is confronted by conflicting data, since the experiments range 
from nearly dry, uncontaminated surfaces on the one hand, 
to completely lubricated surfaces on the other. The following 
generalizations for this condition may be made : 

(a) The coefficient of friction increases with increase of unit 

pressure. 

(b) The coefficient of friction decreases with increase of velocity. 

The value of p. will range between its values for dry, static 
friction and for fluid friction, depending upon the conditions. 
Most machinery bearings fall in this category. Great care should 
be exercised to see that the pressure upon a journal resulting 
from the applied load be not sufficiently great or localized or 
long continued to squeeze out the lubricant already between 
the surfaces and to prevent other lubricant from entering under 
the conditions of speed, etc. Metallic contact, overheating and 
abrasion of the surfaces, even their seizing, may result. When 
a partially lubricated journal is subjected to continuous pressure 
applied at one point in one direction, as for instance, a shaft 

* Kimball, Am. Jour, of Science, 1877; Lanza and Lewis, Trans. A. S. M. E., 
Vol. VII. 



214 MACHINE DESIGN, 

with a constant belt pull or with a heavy fly-wheel upon it, this 
pressure has sufficient time to act and is therefore effective for 
the removal of oil. But if the direction of the pressure is peri- 
odically reversed as at the crank-pin end of a steam-engine con- 
necting-rod, the time of action is less, the tendency to remove the 
oil is reduced, and the oil has opportunity to return between the 
surfaces. Hence a higher unit pressure, p, would be allowable in 
the second case than in the first. If the direction of relative motion 
is also reversed, as at the cross-head pin of a steam-engine, the 
oil not only has an opportunity to return between the surfaces, 
but is assisted in doing so by a sort of pumping action. There- 
fore a still higher unit pressure is allowable. That practical 
experience confirms these conclusions can be seen by reference 
to Table XV. These conclusions are not applicable to the 
cases of forced or flooded lubrication where there is a copious 
supply of oil between the surfaces in any case. 

3. The surfaces are copiously lubricated. — The theory of 
proper lubrication is to provide that the bearing surfaces shall 
be separated always by an unbroken film of lubricant on the 
bearing or pressure side ; there is thus no metallic contact what- 
ever, the journal being fluid-borne. It is obvious how this may 
be done by forced lubrication under sufficient pressure. But 
every continuously rotating journal provided with sufficient oil 
tends to surround itself with an oil-film. The extent to which 
the film is completely formed and maintained will be found to 
depend upon a variety of factors of which the chief are: the 
difference in radius of journal and bore of box, the viscosity of the 
oil, the surface velocity of the journal, the specific load or pres- 
sure per square inch of projected area of journal, and the tem- 
perature of the bearing. 

For the process of film formation see Fig. 122 A. When a 
journal under a given load, P, is at zero velocity it rests on a point 
of contact vertically below its load as seen at a. As rotation 



JOURNALS, BEARINGS, AND LUBRICATION. 



215 



begins, the journal rolls upward on the " on " side of the bearing, 
as shown at b, until the angle <j> included between the line oi 
application of the load and a radial line to the point of con- 
tact equals the " angle of repose " or static friction angle of 
journal and bearing materials when slightly greasy. This is 
the angle whose tangent equals the static coefficient of friction 
of contaminated surfaces as a maximum. When the journal 





Fig. 



'.2A. 



has rolled to this point it begins to slide. If the velocity is very 
low it will continue to slip with contact at this point and the 
ordinary laws of nearly dry friction will govern. As the speed 
increases (the load P remaining constant) the conditions alter. 
Because of its properties of adhesion and surface tension, the 
oil is drawn in by the rotating journal, as illustrated at c, wedging 
the journal completely away from the bearing and moving the 
point of nearest approach over to the " off " side. During 
this period the laws of partial lubrication govern: the coefficient 



216 MACHINE DESIGN. 

of friction falls steadily as the velocity increases, and it is higher 
in value the greater the unit pressure, p, is. 

By this coutinuous action of the lubricant, a film (increasing 
in thickness with increasing velocity of journal surface) is formed, 
separating the journal and bearing. Coincidently, shown in d, 
the point of nearest approach is moved toward point B on the 
horizontal diameter on the " off " side. The pressures in the 
surrounding film are not uniform. A is the point of maximum 
pressure and lies just ahead of the point of nearest approach, 
B, as is to be expected when the wedging action of the oil before 
the narrowed passage at B is considered. Beyond B, at C, is 
the point of minimum film pressure at which experiments show 
that an actual negative pressure or suction exists. 

As the speed increases (the load remaining constant) the 
journal becomes less eccentric and the variations in pressure 
around it also become less. Both A and C move away from B m 
If the speed became infinite the journal would run concentric 
with the bore of the bearing and the points of maximum and 
minimum pressures would be vertically below and above the 
center, respectively. 

If the conditions shown at d had been attained under a 
certain relationship of load and speed and this speed now kept 
constant while the load were increased, the point of nearest ap- 
proach would swing downward again to a position about 40 
from the vertical. With still further increase of load at this 
constant speed the oil film would be ruptured and the conditions 
change again from those of perfect to those of imperfect lubrication. 

But while the condition of complete separation of journal 
and bearing surfaces exists, the friction is fluid friction — the 
correct theory of which was developed independently by Petroff * 
and Reynolds. f In fluid friction: 

* "Neue Theorie der Reibung." Leipzig, 1887. 
f Phil. Trans., 1886. 



JOURNALS, BEARINGS, AND LUBRICATION. 217 

(a) The coefficient of friction varies directly as the viscosity 
of the lubricant. 

(b) The coefficient of friction varies directly as the velocity. 

(c) The coefficient of friction varies inversely as the specific 

pressure. 

(d) The coefficient of friction varies inversely as the mean 

thickness of film on the loaded side of the journal. 

That is, fi = — . 

py 

r) is the " coefficient of viscosity " and may be defined as 
the force necessary to move with unit velocity, one unit of area 
under unit pressure, when the two surfaces are separated by a 
unit thickness of liquid. A study of lubricants shows that f) 
varies inversely as a function of the temperature. On the average 
_ C 

ry ~(/-3*) 2 ' 

V is the journal surface speed; 

p is the pressure per unit of projected area of bearing; 

y is the mean film thickness. 

Strictly this equation should be kept in homogeneous terms; 
practically it is more convenient to express V in feet per minute, 
p in pounds per square inch, y in inches, and rj in pounds per 
square inch for a speed of one foot per minute. 

For the maintenance of a perfect film the relation p = io^V 
may be accepted for speeds from 50 to 500 feet per minute. This 
value is on the safe side according to the experiments of both 
Stribeck * and Tower, f although a little larger than the value 
proposed by Moore J whose experiments were only carried up 
to a speed of 140 feet per minute. 

Stribeck' s results show the breaking-down point of the film 

*Z. d. V. d. I., 1Q02. 

t Proc. Inst. M. E., 1883. 

% Amer. Mach., 1903. 



218 MACHINE DESIGN. 

at ^=20^7. Tower's results on a half-box give />=i2V? 
to 15 VF. The breaking-down point corresponds to the mini- 
mum value of the coefficient of friction. For speeds above 
500 ft. per minute 10VF gives too great a value for p unless 
the bearing is artificially cooled. Above 500 ft. per minute 
P may be taken =30 V y. 

Concerning y the following facts hold : 

(a) It is a function of the running fit allowance. The greater 
the difference in radii of journal and box, a, the larger y has a 
chance of becoming. 

(b) It is a function of p. The relation is complex. On the 
one hand y increases with p because p affects the bore by elastic 
action. An increase of p tends to force out the box at the point 
of nearest approach, thereby increasing the mean film thickness. 
On the other hand y decreases with increase of p, because the 
effect of increase of pressure on the film must be to lessen its 
thickness. The latter effect overbalances the former, hence y 
varies inversely as a function of p. 

(c) It is a function of V. As the velocity increases the film 
builds up and increases in thickness as some function of V. 

(d) It is a function of the temperature, t, of the bearing. 
As the temperature increases and the " body " of the oil dimin- 
ishes in consequence, the film for a given pressure will be re- 
duced in thickness. 

The final result of the combined effects of the various factors 
becomes approximately: 

cv* . . 

* = F^) (I) 

where C is a constant depending upon the running fit allowance 
and the viscosity of the lubricant. For the Deutz engine oil 
used by Stribeck, and a running fit allowance of .001 inch per 
inch of diameter, C=.2i8. 



JOURNALS, BEARINGS, AND LUBRICATION. 219 

To determine n for another lubricant or running fit allow- 
ance it is cnly necessary to vary C inversely as the running fit 
allowance {e.g., double it for a running fit allowance of .0005 
inch per inch of diameter) and to vary it directly as the viscosity 
of the lubricant. Measured with the Engler viscosimeter the 
oil used by Stribeck had a viscosity at 86° F., 20 times that of water 
at 68° F.; and at 104 F., n times that of water at 68° F. 

This equation agrees exactly with Tower's results with sperm 
oil, and very closely with his results on mineral oil,* remembering 
that he has no temperature factor, since his experiments were 
carried out at approximately uniform temperature. It also agrees 
exactly with Hirn's equation f except for this temperature factor. 

The equation has been thoroughly checked on the experi- 
mental data of Stribeck { and holds for speeds from 50 to 500 ft. 
per minute. 

For higher speed bearings it is evident that as the velocity 
increases to infinity y becomes equal to the radial allowance, 
a, irrespective of any finite variation in p. The result would be 
that at very high speeds the coefficient of friction would vary 
directly as the viscosity at the bearing temperature, t, directly 
as the velocity, and inversely as the pressure, 

.*. fJL= — , a being constant. 
pa 

This form seems corroborated by the form of the curves at 
F = 8oo ft. per minute in Stribeck's Figs. 7 and 17. Lasche § 

for high-speed bearings with copious lubrication claims that 

C 

/.'. = for V = 2000, and this is frequently quoted in the 

P(t~3 2 ) 

c.1.2 
form !J.= —f-^- — - for high-speed bearings. But some of Lasche's 

p Pit -3V 

* Thurston, " Friction and Lost Work," p. 313. 
t Unwin, " Machine Design," 1909, p. 232. 
%Z. d. V. d. I., 1902. 
§ " Traction and Transmission," Vol. 6. 



220 MACHINE DESIGN. 

own experiments as shown in his Fig. 26 do not seem to justify 
this conclusion, since they show fJ- still increasing with V at speeds 
approximating 5000 ft. per minute. 

The equation which satisfies his curve 1 (steel journal on 
white metal, ratio of length to diameter 2.2) is found to be 

_ 12^7 

For his curve 4 (nickel steel bearing on gun metal, ratio 
of length to diameter .42), the equation is 

** P(fs*Y 

Since the same oil was used in each case the difference in 
the constants merely points to double the running fit allowance 
in the second case over that of the first. 

Lasche's statement that p. varies as v V up to 500 ft. per 
minute, as v V from 500 to 800 ft. per minute, and is inde- 
pendent of V above 2000 ft. per minute would appear to require 
modification as follows: 

For speeds from 50 to 500 ft. per minute 

V PV -32) 
approximately, 

(C =.2 18 in Stribeck's experiments), 

and above 500 ft. per minute, 

c</v 

"w^ (2) 

approximately, 

(C— 12 in Lasche's experiments.) 

130. Heating of Journals. — To illustrate (4), even if the 
conditions are such that the lubricant is retained between the 



JOURNALS, BEARINGS, AND LUBRICATION. 221 

rubbing surfaces, heating may occur. There is always a fric- 
tional resistance at the surface of the journal; this resistance 
may be reduced (a) by insuring accuracy of form and perfec- 
tion of surface in the journal and its bearings; (b) by insuring 
that the journal and its bearings are in contact, except for the 
film of oil, throughout their entire surface, by means of rigidity 
of framing or self-adjusting boxes, as the case may demand; 
(c) by selecting a suitable lubricant to meet the conditions and 
maintaining the supply to the bearing surfaces. By these 
means the friction may be reduced to a very low value, but it 
cannot be reduced to zero. 

There must be some frictional resistance, and it is always 
converting mechanical energy into heat. This heat raises the 
temperature of the journal and its bearing. If the heat thus 
generated is conducted and radiated away as fast as it is gener- 
ated, the box remains at a constant low temperature. If, how- 
ever, the heat is generated faster than it can be disposed of, 
the temperature of the box rises till its capacity to radiate heat 
is increased by the increased difference of temperature of the 
box and the surrounding air, so that it is able to dispose of 
the heat as fast as it is generated. This temperature, necessary 
to establish the equilibrium of heat generation and disposal, 
may under certain conditions be high enough to destroy the 
lubricant or even to melt out a babbitt-metal box-lining. Sup- 
pose now that a journal is running under certain conditions 
of pressure and surface velocity, and that it remains entirely 
cool. Suppose next that, while all other conditions are kept 
exactly the same, the velocity is increased. All modern experi- 
ments on the friction in journals show that the coefficient of 
friction increases with the increase of velocity of rubbing sur- 

P 2 
face (at speeds above — feet per minute). Therefore the in- 
400 

crease in velocity would increase the frictional resistance at the 



222 MACHINE DESIGN. 

surface of the journal, and the space through which this resist- 
ance acts would be greater in proportion to the increase in 
velocity. The work of the friction at the surface of the journal 
is therefore increased because both the force and the space 
factors are increased. It is this work of friction which has 
been so increased, that produces the heat which tends to raise 
the temperature of the journal and its box. The rate of gen- 
eration of heat has therefore been increased by the increase in 
velocity, but the box has not been changed in any way, and 
therefore its capacity for disposing of heat is the same as it 
was before, and hence the tendency of the journal and its bear- 
ing to heat is greater than it was before the increase in velocity. 
Some change in the proportions of the journal must be made 
in order to keep the tendency to heat the same as it was before 
the increase in velocity. If the diameter of the journal be 
increased, the radiating surface of the box will be proportion- 
ately increased. But the space factor of the friction will be 
increased in the same proportion, and therefore it will be appar- 
ent that this change has not affected the relation of the rate 
of generation of heat to the disposal of it. But if the length 
of the journal be increased, the unit pressure is decreased, which 
tends to decrease the coefficient of friction, while the increase 
of velocity tends to increase it. Due to the combined change 
the coefficient of friction may be slightly increased or decreased, 
the velocity is increased, and the total friction will be slightly 
increased, while the radiating surface of the box is increased 
in greater proportion and the tendency of the box to heat is 
reduced. If the diameter of the journal is reduced coincidently 
with increasing its length, so that the unit pressure remains the 
same both before and after the change, the velocity may be 
reduced, as may also the coefficient of friction, and therefore 
the friction work may be doubly reduced while the radiating 
surface, which will be proportional to the same projected area 



JOURNALS, BEARINGS, AND LUBRICATION. 223 

in both cases, will remain the same. The tendency to heat 
will therefore be reduced. If, therefore, the conditions are such 
that the tendency to heat in a journal, because of the work of the 
friction at its surface, is the vital point in design, it will be clear 
that the ratio of the length of the journal to the diameter is dic- 
tated by it. The reason why high-speed journals have greater 
length in proportion to their diameter than low-speed journals 
will now be apparent. 

If v equals the velocity of journal in feet per second, p being 
the pressure per square inch of projected area and u the coeffi- 
cient of friction, the energy transformed into heat will be upv 
ft. -lbs. per second per square inch of projected area. This 
energy should be dissipated (by radiation and conduction) through 
a surface which bears a relationship to the projected area depend- 
ing upon the thickness of the shell and other portions of the 
bearing. In order to have equilibrium of heat generation and 
heat dissipation a certain temperature, /, in excess of atmospheric 
temperature, / , must be attained. For a ring-oiling bearing 
whose upper radiating surface was about double that of the 
journal under running conditions, Lasche found 

33°° 

and for a bearing with heavy shells of the turbo-dynamo type 
and upper radiating surface about three times that of the journal 

(/-/0 + 33) 2 
^ = — i860— ' 

In both cases / was about 2.5J. 

Stribeck's results on a Sellar's bearing, 1=3-3(1, when similarly 
analyzed show 

2750 



224 MACHINE DESIGN. 

for speeds up to 500 feet per minute, and an apparently more 
rapid radiation still when / -t exceeded 8o°. This checks very 
closely with Lasche's results. 

For a shorter Magnolia metal bearing, l = 2d, Stribeck's 
results give a much greater coefficient of heat dispersion, 

(/-/0 + 22) 2 

f x P v = — i 

225 

but he warns against their reliability. 

It seems safe to assume, where there is no extraneous cooling 
device such as oil or water circulation or fan cooling, that 

&-£*££, ( 3 ) 

2750 

for bearings of ordinary proportions. Thick shells and heavy 
masses in the bearing will increase its heat-radiating capacity 
and thin shells and light masses lower it. 

The nearness of rotating masses acting as air fans will increase 
the heat-dissipating capacity. Dead air spaces around the 
shells decrease it greatly. In case the ordinary proportions of 
the journal give too high a value to /, oil-circulation or even 
water- jacketing systems may be used to dispose of the excess 
heat. A value of t up to 200 is quite safe with ordinary lubri- 
cants.* 

131. Journal Design by Heat Balance. — (a) For velocities 
up to 500 fi. per minute, copious lubrication. 

From equation (1) 

CV* 



From equation (3) 



M = 



Hpv = 



pHt-32)' 

(t-k+35) 2 
2750 ' 



* Dewrance, Proc. Inst. C. E., 1896. 



and 



JOURNALS, BEARINGS, AND LUBRICATION. 225 

. V (/-Zo + 35) 2 

60 2750 
C0 pV (t-to+35) 2 



or, 



/> 5 (/~32) 60 2750 

CV*p*Jt-to + 3 s)*(t-32) 
60 2750 

As previously explained, C=.2i8 for Deutz engine oil and 
a running fit allowance of .001 inch per inch of diameter, 

... F y (t-to+isnt-y) 

IO 

With known values of V and p, and assumed value of t , the room 
temperature, this can be solved by taking trial values of t until 
one is found to satisfy the equation. For other lubricants and 
running fit allowance C should be varied as explained in sec. 129. 

Let ^=ioVV . . . (5) (page 217); 

Hi ■ ■ ■■ (6); 

P = total load on bearing in pounds. If load varies 

throughout cycle, P = mean load during revolution; 

/ = length of journal, inches; 

J = diameter of journal, inches; 

I 
x=~ } assumed, see Table XVI; 
a 

N = revolutions per minute. 

P, N, and x are given. 

'-* » 



«2 26 MACHINE DESIGN. 

From (5), (6), and (7), 



2 
Squaring 

P 2 _ looxdN 
Jd 2 ~ 12 ' 



P bzdN 

Td= 10 ^' 



.0383P 2 

d ~N^> (9) 



Multiplying both sides by rf 2 , 



.o 3 8 3 i» .o38 3 P2 
P AW ' 

P, AT, and a? being known, solve for d. 

From x=-, l=dx. Solve for /. 
d 

P 

From (6) -^=A Determine />. 

From (7) V= . Solve for V. 

12 

Since 

P=ioV*, (5) 

/>* = 3.i67*. 

Substituting in (4) , 

F? (t-t + 3S nt- 32 ) ^ (ii) 

31.6 

Solve this for /. 

If / has too high a value (200 F. is a good maximum value) 
and neither less viscid oil nor greater running fit allowance may 



JOURNALS, BEARINGS, AND LUBRICATION. 



227 



be used, a new value of x giving a greater ratio of length to 
diameter must be tried; or means provided to use extra heavy- 
masses in the shells and bearing or to cool the latter artificially. 
The following table is based on an assumption of t = 68° F. : 



/ 


7* 


V 


78 


2950 


96 


88 


538o 


136 


98 


8850 


180 


108 


13600 


231 


118 


19700 


285 


128 


27400 


344 


138 


36700 


406 


148 


48500 


476* 



* At this point the radiation curve departs from the equation chosen, allowing higher 
values of V. See section 130. 

Having determined p, V, and t, solve for a in equation (i), 

.2i87* 



pHt-32) 

This can be" used to determine the efficiency of the bearing 
which equals: (Total energy received per minute at journal in 
foot-pounds -ptPV)-r- Total energy received per minute at journal 
in foot-pounds. 

The rate of heat generation per square inch of bearing area 
in foot-pounds per second = upv. 

[The foregoing method may also be used in a modified form 
for a given maximum allowable value of /. Given also P, N, 
and x. 

Solve for d, I, p, and 7 as above. 

Compute a from P = pk{t _ 32 y 
Find upv. 

If this exceeds — , extraneous means of cooling 

2 75° 



228 MACHINE DESIGN. 

should be provided, it being assumed that the running fit allowance 
and lubricant may not be changed. 
Two methods may be used : 

(i) Increase the thickness of the bearing (above -) in the 

0-/O+35) 2 

ratio that the excess bears to — . This provides the 

2750 

additional radiating surface necessary. 

(2) Compute, from its specific heat and the permissible 
range of entering and leaving temperature, the quantity of oil 
or water which must be circulated per second through a jacket 
arrangement to carry off the surplus heat generated.] 

(b) For speeds above 500 feet per minute, copious lubrication. 

p = 3 o\^V . . . (12) Page 218. 



P slndN 



(13) 



Cubing, 



P 3 2j,0007ldN 

¥d? = 12 ' 



d* = 



di = 



7069^/3' 

P3 pz 



■ P 7o6oiW 
7 o6 9 iV- 






7069^3 

Solve (14) for d; P, N, and x being given. 
Solve l=dx for /. 



(14) 



JOURNALS, BEARINGS, AND LUBRICATION. 
P 



229 



Id' 



Determine p, from p-- 
From (2), (page 220) 



From (3), (page 224) 



60 2750 

Substituting from (2) 

12F* pV (t-t Q +ss) 2 
P(t~3 2 ) °° 2 75o 

* (/-/o + 35) 2 (^-32)* 



12 


V* 


M -^- 


-3 2 )' 


(<■ 


-<o+35) 2 


fipv 






2750 


J-« 


-<o+3S) 2 



p 



Table, when / = 68°F. 



55o 



t 


v< 


V 


148° 


2800 


572 


158 


3670 


711 


168 


4500 


836 


178 


5600 


997 


188 


6800 


1 164 


198 


8170 


1350 


208 


9800 


1560 



The foregoing method of design was suggested by the papers 
of Mr. Axel Pedersen in. the American Machinist, 1913 and 1914. 

* This equation gives rather higher values to t than will probably be attained 
by bearings of ordinary proportions using customary lubricants. In other words, 
higher values of V than these may be expected in practice at these temperatures. 
The equation is on the safe side. 



230 



MACHINE DESIGN. 



Table XV. — Cylindrical Journal Pressures from Practice 



Kind of Bearing. 



Pressure in lbs. 
per sq. in. of 
projected area. 



Motion intermittent, direction of load reversing, slow speed. 

Crank pins of shearing and punching machines, presses etc, 

Motion an oscillation, direction of load reversing 

Locomotive cross-head pins 

Gas engine cross-head pins 

Air compressor cross-head pins 

Slow-speed stationary engine cross-head pins 

High-speed stationary engine cross-head pins 

Motion a rotation, direction of load changing. 

Locomotive crank pins 

Gas engine crank pins 

Air-compressor crank pins 

Marine engine crank pins 

Slow-speed stationary engine crank, pins 

High-speed stationary engine crank pins (center crank) 

High-speed stationary engine crank pins (side crank) 

Eccentric sheaves 

Motion a rotation, direction of load nearly constant. 

Merchant marine engine, main bearings 

Naval marine engine, main bearings 

Slow-pumping engine, main bearings 

Slow-speed stationary engine, main bearings 

High-speed stationary engine, main bearings 

Gas engine, main bearings 

Air compressor, main bearings 

Car axle journals 

Locomotive and tender axle journals 

Line shafts on bronze or babbitt 

Steel shaft on lignum vitae, water lubrication 

Practice of Gen'l. Elec. Co. J 
Ring-oiling or other copious lubrication. 
Velocity of journal, ft. per minute = V: 

50- 100 

100-2000 

2000-3000 

3000-4000 



3000-7000* 



3000-4000 
2000-3000! 

400-1350 
1000-1860 

910-1675 



1 400- 1 700 
1 000- 2 000 f 
250- 850 
400- 500 
870-1550 
250- 600 
900-1500 
80- 100 



200- 350 
275- 400 

600 
200- 300 
180- 240 
500- 7oof 
150- 250 
300- 600 
400- 550 
100- 150 

35o 



P=l\/V _ 
^=i 5 .6^F 
^=30 a/f_ 

P=4A </V 



* In Vol. 27, Trans. A. S. M. E., pp. 496-497, Mr. Oberlin Smith gives examples 
of journal pressures in presses running as high as 20,000 pounds per square inch 
on hardened steel toggle pins; and 7000 pounds per -square inch, at a surface speed 
of 140 feet per minute, against the cast-iron pitman driving the ram. The journal 
pressure of the main shaft of the second press was 2400 pounds per square inch. 

f Based on maximum explosion pressure. 

J Data from other sources indicate that these values could be increased con- 
siderably with safety. 



JOURNALS, BEARINGS, AND LUBRICATION. 



231 



Departure from his method, based upon different conclusions 
regarding values of ,«, p, etc., are made here, however. 

The method given under (a) may be applied to these bearings 
also, to design for a maximum given value of t. 

132. Allowable Bearing Pressure. — Table XV on p. 230, based 
upon current practice, may be used as a guide by the designer. 
The value to be used in each case is a matter of judgment. The 
allowable pressure depends, among other items, upon the grade 
of workmanship expected as shown in the fit and surface con- 
ditions of the journal and box. 

133- Journal Proportions. — Customary proportions of journals 
may be seen in the following table compiled from current practice: 

TABLE XVI. 



Kind of Journal. 



Value of 



Minimum. Maximum. Average 



Main bearings, marine engines 

Main bearings, center-crank, high-speed engine 
Main bearings, side-crank, slow-speed engine.. 

Main bearings, gas engines 

Crank pins, marine engines 

Crank pins, high-speed engines 

Crank pins, slow-speed engines 

Crank pins, gas engines 

Cross-head pin, high-speed engines 

Cross-head pin, slow-speed engines 

Cross-head pin, gas engines 

Fixed bearings, shafting 

Self-adjusting bearings, shafting 

Generator and motor bearings 

Machine tool bearings 



1 
2 
1-7 



i-5 
3 
2 . 1 

1-5 



i-7 

2 

i-5 

3 

3 

4 

3 

4 



2 2 
1.9 

25 



134. Materials to be Used. — Regarding the materials of 
journals and their boxes the following general statements may 
be made. It must be borne in mind that the terms bab- 
bitt, brass, and bronze cover wide ranges of alloys of varying 
values. 

Cast iron, wrought iron, soft steel, and hard steel will all run 



232 MACHINE DESIGN. 

well at almost any speed on babbitt metal. The pressure per 
square inch which an ordinary babbitt bearing will stand when 
running cool {i.e., at very slow speed), before being squeezed 
out, has been found to be something over 2000 lbs.* 

Cast iron, wrought iron, soft steel, and hard steel will all 
run well on brass and bronze. Brass and bronze of ordinary 
compositions will carry 5000 lbs. per square inch without suffer- 
ing destruction. Bronze, however, is much better than brass. 

Cast iron will run on cast iron where, owing to large bear- 
ing surfaces, the unit pressure is light. Where the pressure and 
speed are high, as in engine-journals, this will not work.f 

In the same way steel will run on cast iron even at high speeds 
if the pressure is light. It has been found that steel will not 
run on cast iron in engine- journals. % 

Wrought iron, soft steel, and hard steel will all run on hard 
steel. 

Steel under steel if hardened and polished will run under as 
high a pressure as 50,000 lbs. per square inch. 

135. Calculation of Journals for Strength. — Journals gener- 
ally form parts of axles on shafts, and the calculation of their 
diameter for strength becomes part of the calculation of the 
shaft. The principles have been developed at length in the 
preceding chapter and need not be repeated here. 

If the journal is so held that it may be considered as sub- 
jected to pure shearing stress, like the crank-pin of a center- 
crank engine, then 

f*A=P, 

in which P = total maximum load ; 

A = total area subjected to stress; 

f s = safe shearing stress for the conditions. 

* C. F. Porter, Trans. A. S. M. E., Vol. Ill, p. 227. 
t Trans. A. S. M. E., Vol. VI, pp. 853-854. 
%Ibid. 



JOURNALS, BEARINGS, AND LUBRICATION. 233 

For a journal subjected to a pure bending moment, 

»-f. 

far 3 

which becomes Pl= for a solid circular shaft. P/=bending 

4 ° 

moment, / = safe unit stress, and r = radius of shaft. This can 

readily be solved for r. 

If the journal be hollow, 

r\ being the external and r 2 the internal diameter. 

For combined bending and twisting such as the main journal 
of a side-crank engine is subjected to, the expression for a solid 
journal is 

TrrS 

/— =o. 35 M 6 +o,6 5 Vikf 6 2 + JH7. 

4 

For a hollow circular section 

^V^ = °-3SM b +o.6 5 VM} +Mf 

M b being the bending moment and M t the twisting moment. 

In general it will be found that journals proportioned for 
strength merely will not have sufficient area to prevent heating, 
so this item must not be overlooked. 

136. Problem. — Design the main journal of a side-crank 
low-speed engine. 
Diameter of cylinder = 16 ins. 
Length of stroke =36 ins. 

Net forward pressure = 100 lbs. per square inch of piston area. 

Suppose the engine capable of carrying full pressure to half- 
stroke. 



234 



MACHINE DESIGN. 



The area of piston =201.06 square inches. 

.*. total net forward pressure = 20,106 lbs. 

At point of maximum torsional effect, which corresponds to 
the position of maximum velocity of piston, no energy is used 
in accelerating reciprocating parts, and 

F p v p =F c v c ; 

F p =net forward force on piston; 
v p = velocity of piston ; 
jF c = force on crank; 
v c = velocity of crank. 

Since v c is less than v P for this position; F c is greater than 



F P , since F ,= 



p"p 



Assuming a connecting-rod length equal to five and a half 
crank lengths gives (Appendix) jF c = 20,500 lbs. 

Since the crank length is 18 inches, and at this position the 
crank and connecting-rod are nearly at a right angle with 
each other, there is a twisting moment at the journal equal to 

Mt = 20, 500X18 = 369,000 inch-lbs. 

There is also a bending moment equal to 20,500 X the dis- 
tance from center of crank-pin to center of main journal. In 

most cases this distance must be 
assumed; for, although the length 
of the crank-pin and the thickness 
of the crank may be known, the 
length of the main journal is un- 
Y known, since this length and the 
journal diameter are the very 
dimensions sought. Assume then 
FlG * I23 ' that the crank-pin is 6 inches long, 

the crank 3 inches thick, and the middle of main journal 6 



i J i 



JOURNALS, BEARINGS, AND LUBRICATION. 235 

inches from the inner face of crank as shown in Fig. 123. 
This will give 12 inches as the lever-arm; .*. the bending moment, 
ilf 6 , =20,500X12 = 246,000 inch-lbs. 

The equivalent bending moment to the combined actual 
bending and twisting moments 

= M eb 

= 0.35 X 246,000 + 0.65V246000 2 + 369000 2 
= 374,375 inch-lbs. 

But M* = ' ; 

4 

. 3 4X374375 
. . r* = : . 

For a main shaft like this / may be taken = 12,000 lbs. 
per square inch for steel. 






374375 



2000 
3.41 inches; 



.*. diameter of journal = 2 X3.41 =6.82, say 7 inches. 

The length according to practice would be about twice this 
diameter,* or 14 inches. This would give a projected area of 
98 square inches and a pressure of something over 200 lbs. 
per square inch of bearing due to steam-pressure alone. 

To get the actual maximum pressure on the journal it 
would be necessary to know the weight of the shaft, flywheel, 
and other attached parts, and properly combine the pressure 
due to these with the pressure due to the steam. 

The rough rule of practice for Corliss engines is to make 
the diameter of main journal equal to one half the diameter 
of the cylinder. 

* See Table XVI, p. 231. 



236 MACHINE DESIGN. 

137. Problem. — Design the crank-pin for the same engine. 
It will be found that the crank-pin must be designed with refer- 
ence to maintaining lubrication, and that it will have an excess 
of strength. 

Allowing 1200 lbs. per square inch of area,* and noting 
from the table that the average practice for this type of engine 
is to make the length of the pin = i.iXthe diameter, f it fol- 
lows that 

I200 

but l=i.id; 

20500 



i.id 2 = 



1200 
and d = 4 inches, nearly; 

.'. /= 1.1 X'4 = 4j inches, say. 

Checking this for strength, considering the pin subjected 

I 
to a bending moment P-, we write 



P 



IJnr* 
2" 4 ' 
P = 20500 lbs., 

-== — = 2.2^ inches; 

2 2° ' 

r = 2 inches, 

/ = stress in outer fiber; 

4X20^00X2.25 . 

.'. / = — ^ =7300 lbs. per square inch; 

71 X O 

which is, of course, a perfectly safe value for wrought iron or 
steel. 

* See Table XV, p. 230. f See Table XVI, p. 231. 



JOURNALS, BEARINGS, AND LUBRICATION. 



237 



138. Problem. — Design the cross-head pin for the same 
engine. This pin also should be designed for maintaining 
lubrication. Allowing 1400 lbs. per square inch as the per- 
missible pressure on the journal,* and noting that the length 
may be taken as 1.3 times the diameter from average practice t 
gives 



dl 



20500 



and 



1400 

l = i.$d } 

, 2 20500 

,\ 1.3d 2 = , 

1400 

d = 3f inches; 

.*. / = 4^ inches. 

Checking this for strength it is evident that the only way 




Fig. 124. 

this pin can fail is by shearing on two surfaces, A-B and D-E 
(see Fig. 124). 



P = 



/a 27ZT 2 J 

20500 
2X7TX2.84 



= 1150 lbs. per square inch. 



This leaves so great a margin of safety that some manu- 
facturers make the cross-head pin of two parts, an inner pin 
of soft, resilient material, sufficiently large to resist the shear- 
ing stress, and an outer hard-steel bushing which surrounds 
the soft pin, but is not allowed to turn on it. The nature of 



* See TableXV, p. 230. 



t See Table XVI, p. 231. 



2 3 8 



MACHINE DESIGN. 



the forces acting on a cross-head pin tend to wear it to an oval 
cross -section. As such wear takes place the bushing can readily 
be given a quarter turn and clamped in the new position. 
(See Fig. 124.) 

139. Thrust-journals. — When a rotating machine part is 
subjected to pressure parallel to the axis of rotation, means 
must be provided for the safe resistance of that pressure. Li 
the case of vertical shafts the pressure is due to the weight 
of the shaft and its attached parts, as the shafts of turbine 
water-wheels that rotate about vertical axes.. In other cases 
the pressure is due to the working force, as the shafts of pro- 
peller-wheels, the spindles of chucking-lathes, etc. The end- 
thrusts of vertical shafts are very often resisted by the a squared - 
up " end of the shaft. This is inserted in a bronze or brass 
"bush," which embraces it to prevent lateral motion, as in 
Fig. 125. If the pressure be too great, the end of the shaft 
may be enlarged so as to increase the bearing surface, thereby 
reducing the pressure per square inch. This enlargement 



r^^ 





Fig. 125. 

must be within narrow limits, however. (See Fig. 126.) AB is 
the axis of rotation, and ACD is the rotating part, its bear- 
ing being enlarged at CD. Let the conditions of wear be con- 
sidered. The velocity of rubbing surface varies from zero 
at the axis to a maximum at C and D. It has been seen that 
the increase of the velocity of rubbing surface increases the 
work of the friction, and therefore the tendency to wear. From 
this it will be seen that the tendency to wear increases from 



JOURNALS, BEARINGS, AND LUBRICATION. 



2 39 



the center to the circumference of this " radial bearing," and 
that, after the bearing has run for a while, the pressure will be 
localized near the center, and heating and abrasion may result. 
Because of low velocity at the center it becomes difficult to main- 
tain the oil film there, which also adds to this local danger. For 
these reasons, where there is a heavy load to be borne, the bearing 
is usually divided up into several parts, the result being what 
is known as a " collar thrust-bearing," as shown in Fig. 127. 




Fig. 127. 

By the increase in the number of collars, the bearing surface 
may be increased without increasing the tendency to unequal 
wear. The radial dimension of the bearing is kept as small 
as is consistent with the other considerations of the design. 
If d„, = mean diameter of collar, its radial width may be made 
I to -j 3 fi - d m ; and the axial thickness, f to J this width. 

It is found that the " tractrix," the curve of constant tan- 
gent, gives the same work of friction, and hence the same ten- 
dency to wear in the direction of the axis of rotation, for all parts 
of the wearing surface. (See " Church's Mechanics," page 181.) 

This has been very incorrectly termed the " anti-friction " 
thrust-bearing. This is far from being the case. The friction 
work for this and all conical thrust-bearings can be shown 
readily to be excessive. Their one advantage is that they are 
easily adjustable. In general they are to be avoided. 

The pressure that is allowable per square inch of projected 
area of bearing surface varies in thrust-bearings with several 



240 MACHINE DESIGN. 

conditions, as it does in journals subjected to pressure at right 
angles to the axis.* Thus, in the pivots of turntables, swing 
bridges, cranes, and the like, the movement is slow and never 
continuous, often being reversed; and also the conditions are 
such that " bath lubrication " may be used, and the allowable 
unit pressure is very high — equal often to 1500 lbs. per square 
inch, and in some cases greatly exceeding that value. 

Tower's investigation of a water-cooled collar bearing showed 
a maximum value of p of 75 lbs. per square inch of net collar 
area at a mean collar speed of 440 ft. per minute, increasing to 
90 lbs. per square inch at a speed of 170 ft. per minute. 

The lowest values of the coefficient of friction, /£, were obtained 
when p=-$vV. At this relation of p to V they ranged from .0286 
to .0348 at p = 6o and 82.5 respectively. When this relation was 
departed from, /x increased in value. When p was 15 and V was 
440, n went up to .0646. Ordinarily collar bearings are limited 
to a pressure of 50 to 60 lbs. per square inch and a coefficient 
of friction of .035 may be expected under customary conditions 
of running. 

If a single collar is used with a mean diameter = d m and a 
radial width = .i$d m , the following equations may be written. 
P= total load, iV = revs. per minute. 

Projected &rea,=7zd m X .i$d m . 

_ P 

V 12 ' 
P 



.47"V 



=wif. 



* See Proc. Inst. M. E., 1888 and 1891, for reports on experiments with 
thrust-bearings. » 



JOURNALS, BEARINGS, AND LUBRICATION. 241 

,222dJ 12 



m 



i ^ 



I.45N 



If there are n collars, 



r 5 p ' 2 



45 Nn? 



For pivot bearings of the general type shown in Fig. 128, 
Reuleaux (Constructor, p. 65) gives for steel on bronze: 

Slow-moving pivots , d = o . 03 5 V P ; 

Up to 150 r.p.m., d = 0.050 v P; 

Above 150 r.p.m., d = 0.004V PN. 

Tower's experiments on pivot bearings of this form under 
conditions of continuous lubrication show that the best results 
are attained when 

p = S.'j / V V, V being the outer circumferential velocity. 



Let d= diameter of 
= N } r.p.m. 


pivot 


sought for a 


load = 


«*, 


and 


a speed 






P- 


P 

4 














P 

nd 2 " 


-"■Wis-. 














4 
















P 2 

~i6~ 


IS-l^dN 
12 , 














242 



MACHINE DESIGN. 



If this relationship of p, V, d, and N obtains and the ar- 
rangement provides for continuous lubrication as shown, even 
if there are no loose rings or disks, a coefficient of friction, jjl, 
as low as 0.005 ma y ^ e expected. This arrangement gives 
film lubrication, but not in its most perfect form. 

The following table may be used as an approximate guide 
in the designing of thrust-bearings. The material of the thrust- 
journal is wrought iron or steel, and the bearing is of bronze 
or brass (babbitt metal is seldom used for this purpose). 



Table XVII. — Thrust Bearings 



Kind of Bearing 


Velocity of Rubbing Surface 


Allowable pressure lbs. 
per sq. in. of net area 
(less oil-grooves, etc.) 






Solid Collar 


100-upward 


50-60 


Flat Pivot, bath lubrication 


Slow and intermittent 


1 000- 1 500 


Flat Pivot, bath lubrication 


100 


ioo*-6oo 


Flat Pivot, bath lubrication 


200 


140* 


Flat Pivot, bath lubrication 


400 


200* 


Flat Pivot, bath lubrication 


800 


280* 


Flat Pivot, bath lubrication 


1600 


400* 


Loose Ring, drill spindle, or 






worm shaft 




336 



* For best efficiency. 

If the journal is of cast iron and runs on bronze or brass, the 
values of allowable pressure given should be divided by two. 

The most efficient forms of thrust-bearings are those * employ- 
ing the principles shown in Fig. 128. 

Between the end of the shaft and the bottom of the step a 
series of accurately finished disks are introduced. The disks are 
alternately hard steel and bronze, the top one is fastened to the 
shaft, the lower to the step, and the rest are free. As indicated, 
each disk has a hole through the middle and radial grooves to 
permit the lubricant to have access between the disks. The 



* See Trans. A. S. M. E., Vol. VI, p. 852, and Proc. Inst. M. E., li 
184; 1891, Plate 30. 



P. 



JOURNALS, BEARINGS, AND LUBRICATION. 



243 



effect of centrifugal force when the shaft is rotating is to force 
the oil outward from between the plates and upward. It is 
collected in the annular chamber a-a and flows from there 
down the drilled passages back to the bottom of the bearing. 
This is equivalent to a continuous automatic pump action sup- 
plying oil to the surfaces. This form of bearing reduces the 
relative motion between successive surfaces to a minimum. A 
similar arrangement of loose disks can be used to great advan- 
tage on small propeller shafts and on worm shafts. 

For thrust-bearings in which the lubricant is automatically 
circulated, or supplied by a force-pump so as to " float " the 




Fig. 128. 



journal, the allowable unit bearing pressures become quite great, 
examples of satisfactory operation at loads as high as 1000 lbs. 
per sq.in. being known. With a lubricant of suitable viscosity 
the conditions, at sufficiently high speeds, would tend to give 
practically fluid friction, i.e., the frictional resistance would be 
independent of the pressure. As the speed decreases the tend- 
ency to maintain the oil- film grows less, however, and there are 
critical speeds corresponding to certain loads at which the film 
appears to break down and seizing takes place. For pivot 
bearings this minimum speed appears to be, from Tower's experi- 



244 



MACHINE DESIGN. 



ments, V=— . Where special means of forced lubrication are 
76 

not employed it will be safe in the design of ordinary thrust-jour- 
nals to use the unit pressures given in Table XVII. 

Fig. 12SA shows the step bearing of the Curtis turbine 




Fig. 12&I. 



employing forced lubrication. The following data * will be found 
of interest in connection with the design of this type of bearing: 



Rating. 


Gals, per min. 


Unbalanced 
weight. 


Step block. 


K.W. 


R.p.m. 


Total. 


i.e. load. 


Outside 
diameter. 


Inside 
diameter. 


750 


1800 


7 


10800 


8 


5i 


1500 


1500 


8 


19700 


11 


8 


3750 


900 


16 


65000 


16 


9k 


5000 


750 


21 


IOIOOO 


16 


10 


9000 


750 


27 


148000 


i8| 


10! 


14000 


750 


32 


190000 


20 


13* 


1500O 


7SO 


36 


216000 


21 


15 


20000 


75° 


43 


233000 


21 


is 



* From Alford's Bearings, McGraw-Hill Co. 



JOURNALS, BEARINGS, AND LUBRICATION. 245 

When bearings have to be used where corrosion or electro- 
lytic action is to be feared, as in turbine work, glass and the 
end grain of very hard woods have been used successfully as 
bearing materials. 

140. Problem. — It is required to design the collar thrust- 
journal that is to receive the propelling pressure from the screw 
of a small yacht. The necessary data are as follows: The 
maximum power delivered to the shaft is 70 H.P. ; pitch of 
screw is 4 feet; slip of screw is 20 per cent; shaft revolves 250 
times per minute; diameter of shaft is 4 inches. 

For every revolution of the screw the yacht moves forward 
a distance = 4 feet less 20 per cent = 3.2 feet, and the speed of 
the yacht in feet per minute = 250X3.2 =800. 

70 H.P. = 70X33,000 = 2,310,000 ft. -lbs. per minute. 

This work may be resolved into its factors of force and space, 
and the propelling force is equal to 2,310,000-^800=2900 lbs., 
nearly. 

The shaft is 4 inches in diameter, and the collars must project 
beyond its surface. Estimate that the mean diameter of the 
rubbing surface is 4.5 inches, then the mean velocity of rubbing 

surface would equal 4.5 X — X 250 = 294 feet per minute. A 

safe value of p, the pressure per square inch, at this speed is 
50 lbs. The necessary area of the journal surface is there- 
fore = 2900 -7-50 = 58 square inches. 

It has been seen that it is desirable to keep the radial dimen- 
sion of the collar surface as small as possible in order to have 
as nearly the same velocity at all parts of the rubbing surface 
as possible. The width of collar in this case will be assumed 
= -^ = 0.75 inch; then the bearing surface in each collar 

5.52 Xtt 4 2 Xtt 

= - — — = 23.7 -12.5 = 11. 2 sq. in. 

4 4 

Then the number of collars equals the total required area 
divided by the area of each collar= 58^11.2 = 5.18, say 6. 



246 



MACHINE DESIGN. 



141. Thrust-bearings with Perfect Film Lubrication. — In 

recent years a great advance has been made in the design of 
thrust-bearings by succeeding in applying to them the principles 




ym 



, 




Position of Shoe in Bearing. 
Fig. 1285. 



of continuous perfect film lubrication. The mathematical analysis 
was first published by Mitchell,* but the earliest practical bear- 

* Zeit. fiir Math, und Physik, 1905. 



JOURNALS, BEARINGS, AND LUBRICATION. 247 

ings designed for use appear to be those of Kingsbury * and 
this type of bearing is known by the latter's name. The bearing 
is submerged in oil. The ring which supports the step or collar 
is not one solid ring, but is divided into segments each one of 
which is pivoted at or near its center of pressure on a spherical- 
ended, cylindrical, upright support. As a consequence, each 
individual segment is free to incline at a small angle (about 
1 3000) with the step surface, allowing the perfect formation 
of the lubricant wedge. 

Steam turbines with these bearings have been run with a 
unit pressure of 500 lbs. per square inch at linear speeds of 3000 
to 4500 ft. per minute, and vertical water wheels with a unit 
pressure of 250 to 400 lbs. per square inch with a coefficient 
of friction generally lying between the remarkably low values 
of .001 and .002. This equals or exceeds ball or roller bearing 
efficiency. Tests show that there is practically no limit of speed 
provided that the oil be circulated and the heat generated in it 
by friction be removed. It will probably be found that the 
allowable pressure will lie in the vicinity of p=2o\/V. These 
bearings have shown great overload capacity under test. 

The center of pressure, for square blocks or those whose 
length is not more than 3 times their width, is about .4 of their 
length from the rear end.f This is the proper location of the 
spherical seat of each segment. 

142. Bearings and Boxes. — The function of a bearing or box 
is to insure that the journal with which it engages shall have 
an accurate motion of rotation or vibration about the given axis. 
It must therefore fit the journal without lost motion; must 
afford means of taking up the lost motion that results neces- 
sarily from wear ; must resist the forces that come upon it 
through the journal, without undue yielding ; must have the 

♦See Alford's "Bearings," p. 150, and Engineering Record, Jan. 11 and 18, 1913. 
t Zeit. fur Math, und Physik, 1905. 



248 



MACHINE DESIGN. 



wearing surface of such material as will run in contact with 
the material of the journal with the least possible friction and 
least tendency to heating and abrasion; and must usually 
include some device for the maintenance of the lubrication. The 
selection of the materials and the providing of sufficient strength 
and stiffness depends upon principles already considered, and 
so it remains to discuss the means for the taking up of necessary 
wear and for providing lubrication. 

Boxes are sometimes made solid rings or shells, the journal 
being inserted endwise. In this case the wear can only be 
taken up by making the engaging surfaces of the box and journal 
conical, and providing for endwise adjustment either of the 
box itself or of the part carrying the journal. Thus, in Fig. 129, 

the collars for the pre- 



venting of end motion 
while running are jam- 
nuts, and looseness be- 
tween the journal and 
box may be taken up by 




Fig. 129. 



Fig. 130. 
moving the journal axially toward the left. 

By far the greater number of boxes, however, are made in 
sections and the lost motion is taken up by moving one or more 
sections toward the axis of rotation. The tendency to wear is 
usually in one direction, and it is sufficient to divide the box 
into halves. Thus, in Fig. 130, the journal rotates about the 
axis O, and all the wear is due to the pressure P acting in the 
direction shown. The wear will therefore be at the bottom of 
the box. It will suffice for the taking up of wear to dress off 
the surfaces at aa, and thus the box-cap may be drawn further 
down by the bolts, and the lost motion is reduced to an admis- 
sible value. "Liners," or "shims," which are thin pieces of 
she,et metal, may be inserted between the surfaces of division 
of the box at aa, and may be removed successively for the lower- 



JOURNALS, BEARINGS, AND LUBRICATION. 



249 



ing of the box-cap as the wear renders it necessary. If the axis 
of the journal must be kept in a constant position, the lower 
half of the box must be capable of being raised. 

Sometimes, as in the case of the box for the main journal of 
a steam-engine shaft, the direction of wear is not constant. 
Thus, in Fig. 131, A represents the main shaft of an engine. 
There is a tendency to wear in the direction B, 
because of the weight of the shaft and its at- 
tached parts ; there is also a tendency to wear 
because of the pressure that comes through the 
connecting-rod and crank. The direction of 
this pressure is continually varying, but the 
average directions on forward and return stroke 
may be represented by C and D. Provision needs to be made, 
therefore, for the taking up of wear in these two directions. If 




>c 



Fig. 131. 




Fig. 



132. 



the box be divided on the line EF, wear will be taken up verti- 
cally and horizontally by reducing the liners. Usually, however, 
in the larger engines the box is divided into four sections, A } B y 



25° 



MACHINE DESIGN. 



C, and D (Fig. 132), and A and C are capable of being moved 
toward the shaft by means of screws or wedges, while D may be 
raised by means of the insertion of " shims. ' ' 

The lost motion between a journal and its box is sometimes 
taken up by making the box as shown in Fig. 133. The exter- 
nal surface of the box is conical and fits in a conical hole in 
the machine frame. The box is split entirely through at A, 
parallel to the axis, and partly through at B and C. The ends 
of the box are threaded, and the nuts E and F are screwed on. 
After the journal has run long enough so that there is an unal- 
lowable amount of lost motion, the nut F is loosened and E 
is screwed up, the effect being to draw the conical box 
further into the conical hole in the machine frame; the hole 




Fig. 133. 

through the box is thereby closed up and lost motion is reduced. 
After this operation the hole cannot be truly cylindrical, and 
if the cylindrical form of the journal has been maintained, it 
will not have a bearing throughout its entire surface. This is 
not usually of very great importance, however, and the form of 
box has the advantage that it holds the axis of the journal in 
a constant position. As far as is possible the box should be 
so designed as to exclude all dust and grit from the bearing 
surfaces. 

All boxes in self-contained machines, like engines or machine 
tools, need to be rigidly supported to prevent the localization 
of pressure, since the parts that carry the journals are made as 
rigid as possible. In line shafts and other parts carrying journals, 



JOURNALS, BEARINGS, AND LUBRICATION. 



251 



when the length is great in comparison to the lateral dimensions, 
some yielding must necessarily occur, and if the boxes were 
rigid, localization of pressure would result. Hence "self- 
adjusting" boxes are used. A point in the axis of rotation at 
the center of the length of the box is held immovable, but the 
box is free to move in any way about this point, and thus adjusts 
itself to any yielding of the shaft. This result is attained as 
shown in Fig. 134. O is the center of the motion of the box; 




Fig. 134. 

B and A are spherical surfaces formed on the box, their center 
being at O. The support for the box contains internal spherical 
surfaces which engage with A and B. Thus the point O is always 
held in a constant position, but the box itself is free to move in 
any way about O as a center. Therefore the box adjusts 
itself within limits to any position of the shaft and hence the 
localization of pressure is impossible. 

In thrust-bearings for vertical shafts the weight of the shaft 
and its attached parts serves to hold the rubbing surfaces in 



252 MACHINE DESIGN. 

contact and the lost motion is taken up by the shaft following 
down as wear occurs. In collar thrust-bearings for horizontal 
shafts the design is such that the bearing for each collar is 
separate and adjustable. The pressure on the different collars 
may thus be equalized.* 

143. Lubrication of Journals. f — The best method of lubrica- 
tion is that in which the rubbing surfaces are constantly sub- 
merged in a bath of lubricating fluid. This method should be 
employed wherever possible if the pressure and surface velocity 
a:e high. Unfortunately it cannot be used in the majority of 
cases. It is not nec:ssa:y that the whole surface be sub- 
merged. If a part of the moving surface runs in the oil-bath it 
is sufficient.! The same result is accomplished by the use of 
chains and rings encircling the journals and dipping into oil- 
pockets, as described later in this section. The effect is to 
form a complete film of oil enveloping the journal. To allow 
this it is evident that the bore of the bearing must be slightly 
greater than the diameter of the journal and a good value to use 
for " running fit allowances " is 0.001 inch per inch of diameter. 
The oil film may be conceived to be made up of a series of 
layers, the one next the bearing surface remaining stationary 
with regard to it, while the layer in immediate contact with the 
shaft rotates with the latter. The intermediate layers, therefore, 
slip upon each other as the shaft rotates and the friction becomes 
very closely akin to ' ' fluid friction " with the bearing floating 

* For complete and varied details of marine thrust-bearings see "Maw's 
Modern Practice in Marine Engineering." 

f See "Lubrication and Lubricants," by Archbutt and Deeley, London. 

| Tower's experiments, Proc. Inst. M. E., 1883 and 1885. See further Prof. 
Reynolds' paper "On the Theory of Lubrication," Phil. Trans., 1886, Part I, 
pp. I57-234- 

§ Professor Reynolds states, in Phil. Trans., 1886, Part 1, p. 161, that if viscosity 
were constant the friction would be inversely proportional to the difference in radii 
of the journal and the bearing. 




< 

<^ 

6 



254 



MACHINE DESIGN. 



upon the lubricant, there being no contact between the metallic 
surfaces. Fig. 134 A shows the conditions of pressure existing 
in the film in Tower's classic experiments. It is impossible 
to introduce oil satisfactorily at the points where the film is 
under pressure; it should be introduced and distributed where 




Fig. 134 B. 



BOTTOM HALF 



the pressure is least. Referring back to Fig. 122^4, d, it will 
be seen that this point is at C, just beyond B, the point of nearest 
approach. Dewrance (Proc. Inst. Civil Engs., 1896) reports as 
high as 30 inches vacuum at this point on a heavily loaded journal. 
Other experimenters confirm this phenomenon. 

Under the action of the load the edges of the boxes tend 
to " pinch in " and scrape off the film from the journal. To 
prevent this these edges should be cut away, thus also forming 
an excellent oil channel for longitudinal distribution of the oil 
where the pressure is least. An excellent arrangement of boxes 
for distributing the oil and maintaining the film is shown in Fig. 
134$, which is copied from Vol. 27, Trans. A. S. M. E., p. 484. 

With pad lubrication or where the oil is fed drop by drop there 
is a tendency for the film to be too thin or to break down, allow- 



JOURNALS, BEARINGS, AND LUBRICATION. 



255 



ing contact of the metallic surfaces, and the highly favorable 
condition of fluid friction disappears. The conditions then lie 
between " fluid friction " and " solid friction " and are too com- 
plex for the statement of consistent results, but it may be approxi- 
mately stated that, with good pad lubrication, the coefficient 
of friction will be about twice that of film lubrication. With 
drop by drop lubrication the value of the coefficient may be- 
come anything between twice that for best film lubrication 
(i.e., =0.0012), and 0.18, the value determined by Morin for 
dry journals. It becomes apparent that some system of forced 
or flooded lubrication whereby 
a continuous film is insured is 
of utmost value in maintaining 
efficiency. 

Let /, Fig. 135, represent a 
journal with its box, and let A , 
B, and C be oil-holes. If oil is 
introduced into the hole A, it 

will tend to flow out from between the rubbing surfaces by the 
shortest way, i.e., it will come out at D. A small amount 
will probably go toward the other end of the box because of 
capillary attraction, but usually none of it will reach the middle 
of the box. If oil be introduced at C, it will come out at E. A 
constant feed, therefore, might be maintained at A and C, and 
yet the middle of the box might run dry. If the oil be introduced 
at B, however, it tends to flow equally in both directions, and 
the entire journal is lubricated. The conclusion follows that 
oil ought, when possible, to be introduced at the middle of the 
length of a cylindrical journal. It should be introduced as far 
as possible from the side where the forces press the journal and 
box closest together.* If a conical journal runs at a high velocity, 
the oil under the influence of centrifugal force tends to go to 



Fig. 135. 



* Tower's experiments, Proc. Inst. M. E., 1883 an d 1885. 



256 



MACHINE DESIGN 



the large end of the cone, and therefore the oil should be intro- 
duced at the small end to insure its distribution over the entire 
journal surface. 

If the end of a vertical thrust- journal whose outline is a 
cone or tractrix, as in Fig. 136, dips into a bath of oil, B, the 
oil will be carried by its centrifugal force, if the velocity be 
high, up between the rubbing surfaces, and will be delivered 
into the groove A A. If holes connect A and B, gravity will 
return the oil to B, and a constant circulation will be main- 
tained. If the thrust- journal has simply a flat end, as in Fig. 
137, the oil should be supplied at the center of the bearing; 
centrifugal force will then distribute it over the entire surface. 
If the oil is forced in under a pressure sufficient to "float" the 
shaft the friction will be greatly reduced. Vertical shaft thrust- 
journals may usually be arranged to run in an oil-bath. Marine 
collar thrust- journals are always arranged to run in an oil-bath. 



A 


1 


A 


9 


§|_ii 


11 


" 


11 



Fig. 136. 





Fig 137. 



Fig. 138 



Sometimes a journal is stationary, and the box rotates 
about it, as in the case of a loose pulley, Fig. 138. If the oil 
is introduced into a tube, as is often done, its centrifugal 
force will carry it away from the rubbing surface unless a " grease 
candle " or other type of pressure lubricating device is used. But 
if a hole is drilled in the axis of the journal, the lubricant intro- 



JOURNALS, BEARINGS, AND LUBRICATION. 



257 



duced into it will be carried to the rubbing surfaces as required. 
If a journal is carried in a rotating part at a considerable dis- 
tance from the axis of rotation, and it requires to be oiled while 
in motion, a channel may be provided from the axis of rota- 
tion, where oil may be introduced conveniently, to the rub- 
bing surfaces, and the oil will be carried out by centrifugal 
force. Thus Fig. 139 shows an engine-crank in section. Oil 
is introduced at b, and centrifugal force carries it through 
the channel provided to a, where it serves to lubricate the rub- 



Fig. 139. 




Fig. 140. 



bing surfaces of the crank-pin and its box. If a journal is 
carried in a reciprocating machine part, and requires to be 
oiled while in motion, the "wick-and wiper " method is one of 
the best. (See Fig. 140.) An ordinary oil-cup with an adjust- 
able feed is mounted in a proper position opposite the end of 
the stroke of the reciprocating part, and a piece of flat wick 
projects from its delivery-tube. A drop of oil runs down and 
hangs suspended at its end. Another oil cup is attached to 
the reciprocating part, which carries a hooked " wiper," C. 
The delivery-tube from C leads to the rubbing surfaces to be 
lubricated. When the reciprocating pait reaches the end of 
its stroke the wiper picks off the drop of oil from the wick 



258 MACHINE DESIGN. 

aiid it runs down into the oil-cup C, and thence to the sur- 
faces to be lubricated. This method applies to the oiling of 
the cross-head pin of a steam-engine. The same method is 
sometimes applied to the crank-pin, but here, through a part 
of the revolution, the tendency of the centrifugal force is to 
force the oil out of the cup, and therefore the plan of oiling 
from the axis is probably preferable. 

When journals are lubricated by feed-oilers, and are so 
located as not to attract attention if the lubrication should fail 
for any reason, " tallow-boxes " or " grease-cups " are used. 
These are cup-like depressions usually cast in the box-cap 
and communicating by means of an oil-hole with the rubbing 
surface. These cups are filled with grease that is solid at 
the ordinary temperature of the box, but if there is the least 
rise in temperature because of the failure of the oil- supply, 
the grease melts and runs to the rubbing surfaces, and sup- 
plies the lubrication temporarily. This safety device is used 
very commonly on line-shaft journals. 

The most common forms of feed-oilers are: I. The oil-cup 
with an adjustable valve that controls the rate of flow. II. The 
oil-cup with a wick feed (Fig. 141). The delivery has a tube 
inserted in it which projects nearly to the top of the cup. In 
this tube a piece of wicking is inserted, and its end dips into 
the oil in the cup. The wick, by capillary attraction, carries 
the oil slowly and continuously over through the tube to the 
rubbing surfaces. III. The cup with a copper rod (Fig. 142J. 
The oil-cup is filled with grease that melts with a very slight 
elevation of temperature, and A is a small copper rod dropped 
into the delivery-tube and resting on the surface of the journal. 
The slight friction between the rod and the journal warms 
the rod and it melts the grease in contact with it, which runs 
down the rod to the rubbing surface. IV. Sometimes a part 
of the surface of the bottom half of the box is cut away and 



JOURNALS, BEARINGS, AND LUBRICATION. 



259 



a felt pad is inserted, its bottom being in contact with an oil- 
bath. This pad rubs against the surface of the journal, is 
kept constantly soaked with oil, and maintains lubrication. 

Ring-and-chain lubrication may be considered as special 
forms of bath lubrication. Fig. 143 shows a ring oiling bearing. 

A loose ring rests on top of the journal, the upper box being 
cut away to permit this; the ring surrounds the lower box 






Fig. 141, 



o 



Fig. 142. 




_. ' ■ ■ ' . 



Fig. 143. 



and extends into a reservoir filled with oil. The rotation of 
the shaft carries the ring with it, which, in turn, brings up a 
constant supply of oil from the reservoir. The annular spaces 
A -A catch all oil which works out along the shaft and return 
ii to the reservoir. 

Modern machines are equipped frequently with complete 
oil distributing and circulating systems, including necessary oil- 
pipes, chambers, cooling and filtering devices and pump. Con- 
tinuous lubrication of this sort, properly applied, leads to very 
high mechanical efficiencies. 

Graphite is winning a deservedly high place as a lubricant for 
certain conditions. Its action is to reduce '' solid friction" by 



260 MACHINE DESIGN. 

filling the inequalities in the surfaces of the relatively moving 
members, giving each a smooth, slippery coating, thereby reduc- 
ing the coefficient of friction. It is particularly useful when the 
conditions of pressure or temperature are such as would tend 
to squeeze out, gum, or destroy liquid lubricants, if these were 
used alone. 

Although it may be applied in some cases in dry flake form, 
it is customary to use it in the form of a mixture with oils, grease, 
or even water. Caution must be observed that the graphite used 
is free from all grit. 



CHAPTER XIII. 



ROLLER- AND BALL-BEARINGS. 



144. General Considerations. — By substituting rolling motion 
in bearings in place of relative sliding, friction losses can be 
greatly reduced. In the design of such bearings there are 
five points to be borne in mind : 

I. The arrangement of the parts and their form must be 
such that their relative motion is true rolling with the least pos- 
sible amount of sliding. This means spheric motion. 

II. The form of the constraining surfaces must be such that 
the rolling parts will not have any effective tendency to leave 
the proper guides or " races." 

III. The rollers and balls must not be unduly loaded. 

IV. Provision must be made to admit the lubricant, and to 
exclude all dust and grit. 

V. The arrangement of the parts must be such as will permit 
unavoidable elastic yielding without causing pinching or binding. 

These points will be considered in the order given. 

145. I. Rolling, Sliding, and Spinning. (See Fig. 144.)— At 




^ 




A is shown the longitudinal section of a cylindrical ball-bearing 
of the simplest form stripped of all auxiliary parts. At B 

261 



262 



MACHINE DESIGN. 



is shown the same for a roller-bearing. At C is a cress-section of 
either, showing but one pair of balls or rollers, R. S is the journal 
and T the box. Consider T as stationary, then the point of 
contact of R and 6* would have the same motion relative to T 
whether considered as a point of R or of S, and if the surface 
friction were sufficient there would be no reason for slippage. 
As a matter of fact, in the actual bearing there will be a slight 
amount of slipping .at both of the points of contact. This form 
of bearing is called the u two-point bearing," because there are 
two points of contact. All cylindrical roller-bearings are of 
this fundamental form. In order to have them of practical use 
the rollers must be held in a case or "cage" so that their axes 
will always remain parallel with the axis of the shaft. Fig. 145 
shows such a "cage" with rollers in place. 




Fig. 145. 

Since the rollers are generally of hardened and ground steel 
the best service with the least wear will be given when the 
engaging surfaces are of the same material. To meet this when 
the shaft is of soft steel, say, and the box of cast iron, a hardened 
and ground-steel ring is fitted over the shaft as a shell and 
another inside the box as a bushing, and the rollers run between 
the outer surface of the former and the inner surface of the 
latter. The necessary room for " play " makes it inevitable that 
the axes of roller bearings will get out of line. If the rollers are 
long enough the resultant forces will break them unless they are 



ROLLER- AND BALL-BEARINGS. 



263 



flexible, i.e., made of helically rolled strips. The alternative 
expedient is that, shown in Fig. 145, of dividing the rollers into 
short, separate lengths. 

Ball-bearings are subject to an action known as " spinning." 
To illustrate this, consider the three-point ball-bearing shown 
in Fig. 146. Here the cehtros are as shown in B, and the con- 
ditions are correct for theoretical rolling as long as point contact 
is maintained and axis C-D remains parallel to axis E-F. But 
when the bearing is in use the points of contact, on each side 
of R, with T become small areas, as shown in B. Considering 
the relative motion of R and T at any instant it will be seen that 




Fig. 146. 



Fig. 147. 



there is an action on each side of the ball akin to that of a small 
thrust-bearing. The rubbing produced in this manner naturally 
causes undesirable friction. This is the action known as " spin- 
ning." It may also be called boring. Spinning or boring may 
be defined as that action of the ball with relation to the con- 
straining surfaces which results from a rotation of the ball about 
an instantaneous axis that is approximately normal (actually, 
anything but tangent) to the constraining surfaces. 

Obviously it is even more marked in the case of a four- 
point bearing, as shown in Fig. 147. 

Here, also, there is pure rolling motion as long as point con- 
tact is maintained, and the axes C-D and E-F remain parallel 
to axis G-H; but as soon as the load is applied the points of 



264 



MACHINE DESIGN. 



contact become areas, and "spinning" results at four surfaces. 
Experiments bear out the conclusion that a properly designed 
two-point bearing will have less friction than a three-point, and 
a three-point will have less than a four-point. 

In a " race " whose radius of curvature is just equal to that 
of the ball the friction becomes excessive. Such races should 
never be used. (See Fig. 148.) They have excessive slippage. 

A force acting at the surface of a ball will tend to rotate 
it about an axis parallel to the tangent plane in which the actu- 
ating force lies; furthermore, this axis will be at a right angle 
with the direction of the force. This is true because it is merely 
a special application of the general law that a force applied to 
a body will tend to move it in the direction of action of the force. 
If other forces, or the form of the constraining surfaces, prevent 
rotation about this axis and cause it to take place about some 
other compromise axis, " sliding " takes place to some extent 
and the efficiency and life of the bearing are lowered. 






Fig. 148. 



Fig. 149. 



The general law for the form of rolling bearings may now be 
stated as follows: 

For true rolling, the constraining surfaces of the journal 
and box (i.e., the "races") must be so formed that the axes of 
relative rotation of the rollers or balls with races and cage will 
all intersect the main axis of the bearing at a fixed point through- 
out the complete revolution of the journal. This may be made 
clear by examples. 



ROLLER- AND BALL-BEARINGS. 



265 



Fig. 149 shows a ball or roller R held between two similar 
plates T and S. The upper plate, T, presses down on R with 
a force P which is transmitted through R to S. 

By the principles of so-called " rolling friction," to roll 
T on R will require a force jP (i.e., proportional to P) to 
overcome the resistance. The motion of T on R causes R 
to roll on 5, to which rolling there is induced a resistance also 
equal to )P, but in the opposite direction as regards R. These 
two forces being equal, opposite, and applied at the same dis- 
tance from the center of R, form a couple whose effect would 
be to give R sl motion of rotation about an axis through its 
center, and perpendicular to the plane in which they both lie. 

This case is similar to those shown in Fig. 144, except that 
in the cases there shown S and T are not plane surfaces. Each 
ball in case A and each roller in case B tends to rotate about an 
axis (relatively to the " cage," not shown) as indicated by the 
dotted lines. In both cases the individual axes all intersect 
the main axis of the journal at a fixed point, namely, at infinity, 
throughout the revolution. The general law for true rolling 
is therefore fulfilled. 

In the cases shown in Figs. 146 and 147, obviously the 
same conditions hold. 

Next consider the thrust- bearings shown in Fig. 150: 





Take case A first. S is the moving member, T the sta- 
tionary member, R one of the balls, and OF is the axis of 
rotation of S relatively to T. The center of the ball is at any 



266 MACHINE DESIGN. 

distance r from the axis O Y, and its points of contact with S 
and T are termed A and B respectively. Relatively to the 
inclosing cage (not shown) all parts of the ball in obedience 
to the acting force tend to rotate about the axis OX, which 
always cuts OF at O. Relative to S, the ball, R, rotates about 
the instantaneous axis OA ; relative to T, about OB. The three 
instantaneous axes of relative motion intersect the main axis 
at O. It is not essential that the angle XOY be a right angle. 
Theoretically the conditions for true rolling are fulfilled. Practically 
there will be boring between the outer end of the ball axis and 
the cage. The greater the load along OY, and the greater the 
angle AOB, the more serious this becomes. 

In case B, as S rotates relative to T, the point D common 
to R and S will have a linear velocity proportional to r 2 and, 
similarly, C's linear velocity will be proportional to Y\. If 
AD and BC were two equal, independent circular disks, each 
would have true rolling motion, and BC would make r± revolu- 
tions, while AD would make r 2 - But BC and AD are both 
disks of the same roller, R, and cannot rotate relative to each 
other; hence they must each make the same number of revo- 
lutions, and points C and D of the disks would have to have 
the same velocity, which is inconsistent with the conditions of 
motions of C and D as points of S. Hence a roller cannot 
be correctly used for a thrust-bearing. Short rollers securely 
held in cages are used in practice, but experiments show that 
they are not as efficient as properly designed forms.* 

Consider case C. Relative to T, the double point D will 
have a linear velocity proportional " to r 2 and C will have a 
linear velocity proportional to r\. Consider AD and BC as 

independent disks so proportioned that j-= =— . If D has a 

* See article by T. Hill in American Machinist, Jan. 5, 1899. Also description 
of bearing by C. R. Pratt, same periodical, June 27, 1901. 



ROLLER- AND BALL-BEARINGS. 267 



linear velocity proportional to r 2 , then the angular velocity 
of AD about its axis OX will be proportional to — ry;. 
Similarly, the angular velocity of BC about axis OX will be 



proportional to „,. 






Angular velocity of AD 
Angular velocity of BC 


r 2 

71 AD 


BC r 2 f\ r 2 
AD ri r 2 n * 



nBC 

bc r x 

since "TH =— • 

. AD r 2 

Hence the disks AD and BC have the same angular velocity 
about the axis OX, and may form parts of the same body. 
This will be true of any pair of disks of the cone OBC. Any 
frustum cf a cone whose apex lies anywhere on the axis OY 
will therefore fulfill the conditions for true rolling motion rela- 
tively to T when actuated by S. 

In each of the foregoing cases the rolling members must be 
held in suitable " cages," or they will yield to the tendency to 
displace them. 

In ball thrust-bearings it is desirable to so arrange the balls 
in the cage that each one will have a separate path, as this 
minimizes wear. 

For a three-point thrust ball-bearing the form of the races 
to permit true rolling must be as shown in Fig. 151 to be in 
accordance with the principles just demonstrated. The groove- 
angle should be as flat as possible to reduce the friction effect 
of " spinning." 

About 120 will be found a good practicable value. 

The ball becomes akin to a cone as far as its relations with 
T and S are concerned. In each case the motion imparted to 



268 



MACHINE DESIGN. 



the ball tends to rotate it about the correct axis OX and the con- 
ditions for true rolling are satisfied. The sides of the race are 
tangent to the ball where it is cut by any line A-B which passes 
through O. Boring at A and B is unavoidable. 

A four-point ball-bearing must be designed according to 
the principles indicated in Fig. 152 for true rolling motion. As 





Fig. 152. 



Fig. 153. 



far as its motion relatious with T and 5 are concerned, the ball 
becomes akin to a cone.* Boring occurs at all four tangent points. 

Similarly a cup and cone three-point bearing should have 
the form shown in Fig. 153. 

146. II. The form of the constraining surfaces must be such 
in ball-bearings that the balls will not have any effective tendency 
to leave their proper paths. The use of cages for this purpose 
has already been mentioned. 

* The method of laying out the groove in Fig. 151 is as follows: — The axes of 
rotation of the balls cut the main axis of the bearing at O. Draw the lower surface 
of .S tangent to the ball at C and parallel to the ball axis OX. Draw the line OB 
cutting the ball at A and B, and draw tangent surfaces normal to the radii of the 
ball at A and B. These surfaces form the groove angle BDA. If the first trial 
gives too sharp a groove angle, increase the angle XOB and repeat the construction. 
If BDA is too flat, decrease XOB. 

For the four-point bearing shown in Fig. 152 the same method is used for deter- 
mining the groove in 5 as well as T. 



ROLLER- AND BALL-BEARINGS. 



269 



If two-point bearings without cages are desired, the section 
of each race should be the arc of a circle whose radius is -A 

1 o 

to } of the diameter of the ball. In two-point bearings the 
points of pressure must always be diametrically opposite except 
as noted for thrust-bearings. 

In three- and four-point bearings where the races are properly 
formed for true rolling, as explained in the preceding section, the 
tendency for the balls to leave the races is reduced to a minimum. 

One point, however, needs further consideration. In cup- 
and cone-bearings it is impossible to 
keep a tight adjustment at all times, 
and the least play will allow some of 
the balls on the unloaded side of the 
bearing to get out of place. 

Fig. 154 shows such a bearing 
loosely adjusted. 

The loaded cup is forced down so 
that its axis lies below the axis of 
the cone. The top ball is held correctly in place for true 
rolling; the lower ball is free to roll to one side as seen. 
Investigation has shown that the angles a and /? should each 
be at least as great as 25 in order to return the displaced ball 
easily to its proper path by the time it becomes subjected to 
the load. If the angles are too acute there is a tendency for 
the balls to wedge in their incorrect positions, causing rapid 
wear or even crushing.* 

147 III. Allowable Loading. — Careful experiments show 
that for high efficiency and durability the loads on balls and 
rollers should be very much less than they could be with safety 
as far as their strength is concerned. f 




Fig. 154. 



* See article by R. Janney in American Machinist, Jan. 5, 1899. 

f See excellent article by Professor Stribeck in Z. d. V. d. I., Jan. 19 and 26, 



1901, 



270 MACHINE DESIGN. 

Let P equal the load in pounds which is allowable for a single 
ball or roller. Then for balls 

d = diameter of ball in inches. 

K = 1 500 for hardened steel balls and races, two-point bearing, 

with circular-arc races having radii equal to § d. 
K= 750 for hardened steel balls and races, three- and four- 
point bearing. 
For two-point bearing with flat races, K = $oo. 
For cast-iron balls and races use two-fifths of these values. 
For bearings in which the greatest care has been taken re- 
garding the selection of the most suitable steel, its proper heat 
treatment, and accuracy of workmanship, these values may be 
increased 50 per cent. 
For rollers P = Kdl. 

d = diameter of roller in inches = mean diameter of cone, 
/ = length of roller in inches, 
K = 4oo for cast iron, 
K= 1000 for hardened steel. 
In thrust-bearings, if the total load = P and the number of 

P 

balls =n, we have for either balls or rollers Po = — . 

n 

In cylindrical bearings the load is always greatest at one 

side of the bearing, the balls or rollers on the opposite side 

being entirely unloaded. It has been found that the load on 

the heaviest loaded ball or roller = P = — P, where n is the 

n 

number of balls.* 



* Mr. Henry Hess, of the Hess-Bright Mfg. Co., in a letter to the authors says: — 
"It is a fact, that has been determined by experience, that in radial [i.e., cylindrical] 
ball bearings the speed has very little influence within very wide limits. In my 
practice ... I pay no attention to speed of radial bearings up to 3000 rpm as 



ROLLER- AND BALL-BEARINGS. 



27I 



For radial bearings, not subjected to shock, speed up to 3000 
r.p.m. need not affect the allowable load. 
For thrust ball-bearings, 

$ooond 2 



v r.p.m. 

for best material and workmanship and circular arc races. For 
cast iron or soft steel divide by 100. 

148. Size of Bearing. — To determine the size of the ball 



influencing the load so long as such speed is fairly uniform and so long as the load 
is fairly uniform. When neither speed nor load are uniform the percussive effect 
of rapid changes must be taken in consideration; unfortunately, so far at least, the 
factors are entirely empirical and allowances are made by a comparison with 
analogous cases of previous practice. 

The case is different with thrust bearings. In these, speed is a very decided 
factor in the carrying capacity even though speed and load be uniform. Here 
again no rational formula has yet been developed to adequately represent the 
different elements, but carrying capacities for different speeds of standard bearings 
have been experimentally determined, since it was quite feasible to get different 
uniform speeds and determine under what load the carrying capacity was reached. 
We found, for instance, that for a thrust bearing employing 18 — \" balls, the 
permissible load at 10 rpm was 2400 pounds; at 300 rpm — 650 pounds; at 1000 — 
450 pounds; and at 1500 only 330 pounds. We also found that, generally speaking, 
it was inadvisable to use this type of bearing for speeds materially above 1500 rpm." 

An analysis of certain standard thrust bearings in connection with the makers' 
catalog allowances for loads, gives at various speeds : — 



Load per Ball, Pounds. 



R. P. M. 


1/4" Ball. 


5/16" Ball. 


3/8" Ball. 


1500 


27.8 


35-4-44-7 


80.7 


1000 


33-3 


4I-7-57-I 


91.7 


500 


41.7 


52.1-71.4 


129 


300 


55.6 


66.7-89.2 


153 


150 


61.2 


8 3-3-io7 


193 


TO 


210 


229-339 


560 



C. R. Pratt, Trans. A.S.M.E., Vol. 27 gives as limiting load per \" ball, 100 
pounds at 700 rpm with a 6" diameter circle of rotation. 



27: 



MACHINE DESIGN. 



circle (i.e. the middle diameter of the u race ") given the 
number of balls n and their diameter d. (See Fig. 155.) 

r= radius of ball. R = radius of ball 
circle. Join the centers of two consec- 
utive balls by the chord AB = 2r. From 
the center of the ball circle, O, draw two 
radii, one to A and the other to the mid- 
point of A-B. Call the angle included 
between the radii a. Then 




Fig. 155. 



since 



r = R sin a i 

180 



and, 



R = 



180 ' 



sin 



This is the radius of a circle on which the centers of the 
balls will lie when their surfaces are all in contact. It is desir- 
able to allow some clearance between the balls. This may- 
be as much as 0.005 mcn between each pair of balls provided 

d 



the total allowance does not exceed 



When the total clear- 



ance has been decided upon, it may be allowed for by making 
the actual radius of the ball circle larger than R by an amount 
one sixth of the total clearance desired. 

The most satisfactory service seems to be given by those 
radial bearings which employ some type of elastic ball sep- 
arators. 

149. IV. Lubrication and Sealing. — On account of " spin- 
ning," faulty adjustment, and unavoidable slippage, rolling 
bearings should be properly lubricated. As they are extremely 
sensitive to the presence of dust and grit, care must be exer- 



ROLLER- AND BALL-BEARINGS. 273 

cised that the lubricant be admitted without any danger of the 
entrance of these. 

Sealed oil-holes, dust-caps, and felt washers are commonly 
used both to retain the lubricant for bath lubrication and for 
keeping out all dirt. 

150. V. Prevention of Binding. — Many ball bearing instal- 
lations, otherwise properly designed, have failed because they 
neglected to take into account the fact that no material is utterly 
rigid or workmanship mathematically accurate. For these 
reasons a bearing employing a double row of balls so arranged 
as to ignore accommodation to natural elastic yielding will not 
carry twice the load of a similar bearing with one row of balls. 
Binding will inevitably result where no provisions for elastic 
yielding have been made. The accommodation to temperature 
changes must also be considered in some installations. 

Manufacturers issue valuable data sheets upon these matters 
and invite consultation as to details of installation. 

151. Efficiency of Ball- and Roller-bearings. — Although the 
efficiency of these bearings, as shown by Stribeck,* Thomas, f 
and others, shows a variation with changes of load, velocity, and 
temperature, these variations are within a relatively small range 
for properly designed and installed bearings. The coefficient 

Fr 

of friction, referred to the force at the shaft = 11= — , where Pr= the 

Pr 

turning moment exerted on the shaft and Fr the corresponding 
friction moment, r being the shaft radius. 

For roller bearings a ranges in value from 0.0035 to °-° 2 > 
the higher value corresponding to great underloading. For 
roller bearings properly proportioned to their load, a mean value 
of fi = 0.005 mav b e used. 

For ball bearings, properly proportioned and correctly in- 
stalled, a mean value of p= 0.002 may be used. 

* Z. d. V. d. I., 1901 and 1902. t Trans. A. S. M. E., 1913. 



CHAPTER XIV. 



COUPLINGS AND CLUTCHES. 



152. Couplings and Clutches Defined. — Couplings are those 
machine parts which are used to connect the ends of two shafts 
or spindles in such a manner that rotation of the one will pro- 
duce an identical rotation of the other. They are therefore 
in the nature of fastenings, and may be classified as perma- 
nent or disengaging. The latter are frequently called clutches. 
- 153. Permanent Couplings. — The simplest form of per- 
manent coupling is shown in Fig. 156, and is known as the 
" sleeve " or " muff " coupling. Each shaft has a keyway cut 




Fig. 156. 



at the end. The cast-iron sleeve of the proportions indicated is 
bored an exact fit for the shafts and has a keyway cut its entire 
length. When the sleeve is slipped over the ends of the shafts, 
the key is driven home and all relative rotation is prevented. 
The key may be proportioned according to the rules laid down 
in Chapter IX. 

274 



COUPLINGS AND CLUTCHES. 



275 



154. Flange couplings are frequently used, and Fig. 157 
illustrates the type. Approximate proportions are indicated. 




Fig. 157. 

The number of bolts n may be from 3 + 0.5^ to 3+d. Their 
diameter d' must be such that their combined strength to resist 
a torsional moment about the axis of the shaft will be equal 
to the torsional strength of the shaft, 



xd 3 n nd'* tt 



/• = allowable stress in outer fiber of shaft, pounds per 

square inch; 
d = diameter of shaft, inches; 

ic = radius of bolt circle, inches; approximately i.$d; 
n= number of bolts; usually an even number, 4, 6, 8, etc.; 
d' = diameter of bolts, inches; 

/e' = allowable shearing stress in bolts, pounds per square inch. 
This equation will approximately give 



d' = 






155. Compression couplings of three forms are shown in 
Figs. 158, 159, and 160. The first is similar to the ordinary 



276 



MACHINE DESIGN. 



flange coupling except that the flanges draw up on a sleeve which 
is split in halves longitudinally and is tapered toward each end 




Fig. 158. 

on the outside. The two flanges have internal tapered surfaces 
to fit these. 



Tit Key 




Fig. 159. 

Instead of being held by rings the half sleeves are sometimes 
bolted together as shown in Fig. 159. 




Section a-b 




Fig. 160. 



Fig. 160 A. 



156. The "Sellers" coupling is shown in Fig. 160. An 
outer sleeve A is bored tapering from each end. A split cone 
bushing B is inserted at each end. Openings are left for three 



BELTS, ROPES, BRAKES, AND CHAINS. 



297 



cds, the centrifugal force; 

pds, the pressure between the face of the pulley and ds; 

dF = ptpds, the friction between the element of belt and pulley 
face. 

These correspond to any cross-sectional area, A, square 
inches. 




Tt is more convenient to develop the equations if a cross- 
sectional area of one square inch is considered. For such a 

T + dT^ 



section the forces acting become *=( — ). 
pds, and ;ipds. 



t + dt = 



-J, cds, 



278 



MACHINE DESIGN. 



M-5 



a 



the reader is referred to Professor Kennedy's " Mechanics of 
Machinery." 

159. Hall's Coupling.* — A novel coupling for transmitting 
motion from axes at 90 at equal, uniform angular velocity 
(equivalent to a pair of miter gears) is the invention of C. P. 
Hall. As seen in Fig. 162^, the coupling heads are merely cyl- 
inders, bored for and keyed to their respective shafts. Near 

their circumference and evenly 
spaced, holes are drilled to receive 
comfortably a number of rods. 
These rods are bent to exactly 90 
and each leg has a length of a coup- 
ling head plus the shortest exposed 
length as shown at B. They are 
free to turn in their sockets and to 
slide lengthwise as the relative 
movement of the heads demands. 
When in the extreme position A, 
the ends of a rod are midway in the heads, and in position B 
are flush with the outer faces. The device should be enclosed 
in a case and well lubricated. The area of the rods should be 
made sufficient to equalize their carrying power to the torsional 
strength of the shaft. 

160. Flexible Couplings. — Where shafts which are or which 
may become slightly out of alignment are to be connected, 
some form of flexible coupling is advisable. Their principle 
is illustrated in Fig. 163. Each shaft has keyed to its end a 
disk which has set in its face a number of pins. The pins are 
so placed that those in the one circle will not strike those in 
the other if either shaft is rotated while the other remains at 
rest. When one shaft is to drive the other, short belts are 



Fig. 162^. 



* Power, Jan. 26, 1915. 



COUPLINGS AND CLUTCHES. 



279 



placed on the pins as shown in B, Fig. 163. The same general 
idea is used in a coupling employing a single continuous 
belt. 

Another device for the same purpose is one which employs 
a flexible disk, shown in Fig. 163^4. 




Fig. 163. 



Fig. 163,4. 



161. Disengaging couplings are of two general classes: 
positive drive and friction drive. Positive-drive couplings are 
commonly called toothed or claw couplings. They consist 
of two members having projections on their faces, as shown 
in Fig. 164, which interlock when in action. A is keyed rigidly 




Fig. 164. 



to its shaft. B can slide along its shaft guided by the feather, F, 
but cannot rotate except with the shaft. When B is moved 



300 MACHINE DESIGN. 

For speeds below 1800 ft. per minute equation (3) may be 
written as 

h 

log T = 4343^ (30 

h 

To apply these equations it is simplest to decide upon a max- 
imum value of h. This varies with the quality of the belt, the 
nature of the splice, etc., and may be taken as high as 300 lbs. 
per square inch; but when the economic life of the belt is con- 
sidered, 200 to 240 lbs. per square inch is better. For a new 
belt take 240 lbs. 

It is evident that the foregoing equations may also be written 
in the following forms : 

T 1 -T 2 = P J (w) 

Vt 1 -Vt 2 = 2 Vt , ( 2 a) 

T —T 
log ,/ J = -4343/^> (3<*) 

and for low velocities 

T 

log ^r=. 4343^ (3<0 

The weight of ordinary oak-tanned leather belting per cubic 
inch may be taken, w = 0.035. 

Table XIX. 

■XT 1 C * T-2WV 2 

Values of t c = . 



V 


1800 


2400 


3000 


3600 


4200 


4800 


5400 


6000 


6600 


7200 


7800 


V 


30 


40 


50 


60 


70 


80 


90 


100 


no 


120 


130 


tc 


11. 7 


20.8 


32.6 


47.0 


64.0 


83.2 


106 


130-5 


158 


188 


221 



From Table XIX it becomes evident that the centrifugal 
tension, which diminishes the effective tension-producing pressure 
between belt and pulley upon which the frictional driving power 



BELTS, ROPES, BRAKES, AND CHAINS. 



2 93 



made in the variables R and r, so long as their sum is constant, 
will not affect the value of the equation, and hence the belt 
length will be constant. It will now be easy to design cone 
pulleys for a crossed belt. Suppose a pair of steps given to 
transmit a certain velocity ratio. It is required to find a pair 
of steps that will transmit some other velocity ratio, the length 
of belt being the same in both cases. Let R and r = radii of the 
given steps; R' and r' = radii of the required steps; i£ + r = 
R'+r'=a; the velocity ratio of R f to f =b. There are two 




Fig. 177. 

equations between R r and r' ', R' -^r f =b, and R' + / =a. Com- 
bining and solving, it is found that r' =a-r-(i +b), and R' =a — r'. 
For an open belt the formula for length, using same symbols 
as for crossed belt, is 



L = 2V&-(R-rf+n{R+r) 

-\-2(R — r) (arc whose sine is (R — r) + d). 

If R and r are changed as before (i.e., R-\-r = R' + r f =ci 
constant), the term R—r would of course not be constant, and 
two of the terms of the equation would vary in value; therefore 
the length of the belt would vary. The determination of cone 
steps for open belts therefore becomes a more difficult matter, 
and approximate methods are almost invariably used. 

172. Graphical Method for Cone-pulley Design. — The fol- 
lowing graphical approximate method is due to Mr. C. A. Smith, 
and is given, with full discussion of the subject, in "Transactions 



282 MACHINE DESIGN. 

Table XVIII. 

lx=o.io to 0.15 for cast iron on cast iron 
= 0.15 to 0.20 for cast iron on paper 
= 0.20 to 0.30 for cast iron on leather 
= 0.20 to 0.40 for cast iron on wood 
=0.33 to 0.37 for cast iron on cork. 

The range in values of n is due, not only to surface differences, 
or the presence of various amounts of lubricating matter, but 
also to variety in the rate of slippage. With metal on metal /* 
decreases, from its value as the coefficient of friction of rest, 
with increase of velocity of slip. But the reverse is true with 
leather. 

It is obvious, in the elementary cone clutch of Fig. 165, that 
the end thrust P produces an undesirable and excessive thrust- 





Fig. 166 A. 



Fig. i665. 



bearing friction on each shaft. For this reason the self-sustaining 
principle is used. The load may be applied by a spring as 
in Fig. i66^4, or by an adjustable self-locking thrust-block device, 
as in Fig. i66jE>, but in either case there is no external end-thrust 
when the clutch is driving. 



COUPLINGS AND CLUTCHES. 



283 



163. Weston Friction Coupling. — For heavy duty the prin- 
ciple of the Weston friction coupling, as shown in Fig. 167, may 
be used. The sleeve A carries two feathers on which a number 




of iron rings C can slide but not rotate. Similarly the hollow 
sleeve B is provided with feathers which prevent the rotation 
of the wooden rings D, while not interfering with their sliding. 




Fig. 167 A, 



Let there be n iron rings. Then, when B is pressed toward 
A t there will be friction induced on 271 + 1 surfaces. If P is 
the axial pressure and //. the coefficient of friction, the total 
friction F = fiP(2n + i), and if r = the mean radius of the rings, 
the moment which can be transmitted = M = Fr. 

This clutch belones to the class known as disk clutches. They 



296 



MACHINE DESIGN. 



this equality continues; i.e., as long as Ti -T 2 = P, in which 
Ti = tension in the driving side and T 2 = tension in the slack 
side. (See Fig. 180.) The tension in the driving side is increased 
at the expense of that in the slack side, but the sum does not 
remain a constant. Analysis of experimental data * shows that 
a close approximation is given by the simple equation 

V7\+V t T 2 = 2Vt . 

T 

To find the value of — . The increase in tension from the 
J- 2 

slack side to the driving side is possible because of the frictional 

resistance between the belt and pulley surface. Consider any 



Fig. 179. 






Fig. ii 



element of the belt, ds, Fig. 181. It is in equilibrium undei 
the action of the following forces: 

T, the value of the varying tension at one end of ds; 

T + dT, the value of the varying tension at the other end of ds 

* Wilfred Lewis in Trans. A. S. M, E„ Vol, VII, 



BELTS, ROPES, BRAKES, AND CHAINS. 305 

of 100-^0.433=231 feet. The work done pe. minute in pump- 
ing the water therefore is equal to 908 lbs. X 231 ftet =209,748 
ft.-lbs. The velocity of the rim of the belt-pulley is equal to 
300 xi. 5 X~= 1414 feet per minute.* Therefore the force P - T\ 
— 7^2 = 209,748 ft.-lbs. per minute -m 41 4 feet per minute = 
148 lbs. 



Fig. 182. 

To find a (see Fig. 182) sin 8 — * T = - = 0.0625. 

I 144 

Therefore ^ = 3° 3 5'; a = i8o -2/? = i8o°-7° 10' = 172° 50'; a 

in n measure = 1721X0.0175 =3.025 =6. 

T 
log ;=- = 0.4343X^X6'= 0.4343X0.3X3.02 5 =0.3941. 

J- 2 

... ^ = 2.48; p = r 1 -r 2 = i 4 8. 

Combining these equations T\ is found to be equal to 248 lbs., 
the maximum stress in the belt. 

T x 248 
The cross-sectional area of belt should be equal to — = — 

t\ 240 

— T.03 square inch. 

Single-thickness belting varies from 0.2 to 0.25 of an inch in 
thickness, hence the width called for by our problem would be 

I -°3 -i 

=5 inches, say. 

0.2 

176. Problem. — A sixty-horse-power dynamo is to run 1500 

revolutions per minute and has a 15-inch pulley on its shaft. 

* At this speed the simple form of the belt formula may be used. 



CHAPTER XV. 

BELTS, ROPES, BRAKES, AND CHAINS. 

165. Transmission of Motion by Belts. — In Fig. 169, let A 
and B be two cylindrical surfaces, free to rotate about their axes; 
let CD be their common tangent, and let it represent an inex- 
tensible connection between the two cylinders. Since it is 
inextensible, the points D and C, and hence the surfaces of the 




cylinders, must have the same linear velocity when A is rotated 
as indicated by the arrow. Two points having the same linear 
velocity, and different radii, have angular velocities which are 
inversely proportional to their radii. Hence, since the surfaces 
of the cylinders have the same linear velocity, their angular 
velocities are inversely proportional to their radii. This is true 
of all cylinders connected by inextensible connectors. Suppose 
the cylinders to become pulleys, and the tangent line to become 
a belt. Let CD' be drawn; this becomes a part of the belt 
together with the portions DED' and CFC, making it endless, 
and rotation may be continuous. The belt will remain always 
tangent to the pulleys, and will transmit such rotation that the 

286 



BELTS, ROPES, BRAKES, AND CHAINS. 287 

angular velocity ratio will constantly be the inverse ratio of the 
radii of the pulleys. 

The case considered corresponds to a crossed belt, but the 
same reasoning applies to an open belt. (See Fig. 170.J A and 




B are two pulleys, and CDD'C'C is an open belt. Since the 
points C and D are connected by a belt that is practically inex- 
tensible, the linear velocity of C and D is the same; therefore 
the angular velocities of the pulleys are to each other inversely 
as their radii. If the pulleys in either case were pitch cylinders 
of gears the condition of velocity would be the same. In the 
first case, however, the direction of motion is reversed, while 
in the second case it is not. Hence the first corresponds to 
gears meshing directly with each other, while the second corre- 
sponds to the case of gears connected by an idler, or to the case 
of an annular gear and pinion. While in many places positive 
driving-gears are indispensable, it is frequently the case that the 
relative position of the axes to be connected is such as would 
demand gears of inconvenient or impossible proportions, and 
belts are used with the sacrifice of positive driving. 

Of course it is necessary that a belt should have some thick- 
ness; and, since the center of pull is the center of the belt, it 
is necessary to add to the radius of the pulley half the thickness 
of the belt. The motion communicated by means of belting, 
however, does not need to be absolutely correct, and therefore 
in practice it is usually customary to neglect the thickness of the 



308 MACHINE DESIGN. 

and decrease \± in the expression fipds = dT, and the converse 
is also true. This is probably of no practical importance. 

The value of P may also be increased by increasing either /*, 
the coefficient of friction, or 6, the arc of contact, since increase 

T 1 

of either increases the ratio — , and therefore increases 

J- 2 

T 1 -T 2 =P. 

Increasing To decreases the life of the belt. It also in- 
creases the pressure on the bearings in which the pulley-shaft 
runs, and therefore increases frictional resistance; hence a 
greater amount of the energy supplied is converted into heat and 
lost to any useful purpose. But if T is kept constant, and /x 
or is increased, the driving power is increased without noticeable 
change of pressure in the bearings, since Vj\+Vt 2 remains 
constant. When possible, therefore, it is preferable to increase 
P by increase of p. or 6, rather than by increase of T . 

Application of belt-dressing may serve sometimes to in- 
crease fi. 

If, as in Fig. 179, the arrangement is such that the upper 
side of the belt is the slack side, the "sag " of the belt tends to 

T 

increase the arc of contact, and therefore to increase 7^-. If 

the lower side is the slack side, the belt sags away from the pulleys 

T 
and d and ttt are decreased. 
-L 2 




Fig 183. 
An idler-pulley, C, may be used, as in Fig. 183. It is pressed 
against the belt by some means. Its purpose may be to increase 

/ -\J nr 1 ■y/'T \ 

P by increasing the tension, T =[ I I} 2 . In this case 



BELTS, ROPES, BRAKES, AND CHAINS. 301 

depends, increases rapidly with the velocity and if t\ = 240 lbs. 
per square inch there will be no effective pressure at a speed of 
about 8000 ft. per minute. In other words the belt, if put on 
with the proper value of / corresponding to 21 = 240, can trans- 
mit no power at this speed, because the centrifugal force is so 
great that no pressure exists between the belt and the face of 
the pulley, and hence there is no friction. 

The necessary value of P is a given condition in any problem. 
If the power to be transmitted by the belt is given in HP and 
the velocity of the belt, V, in feet per minute is known, 

HP X 33,000 



P = 



V 



The most economical speed at which to use a leather belt is 
about 4500 ft. per minute. In general P is determined by 
dividing the foot-pounds of work per minute to be transmitted, 
by the belt speed (or pulley rim velocity) in feet per minute. 

The value of is determined for the pulley of smallest arc 
of contact from the diameters of the pulleys and the distance 
between their centers. (See sec. 174.) 

The value of fi, the coefficient of friction, varies with the 
kind of belting, the material and character of surface of pulley, 
the condition of the belt as regards dressing, the side of the 
leather used, and particularly with the rate of slip of the belt 
on the pulley. This slip is a compound of two factors, actual 
slippage and belt creep, the latter being the unavoidable move- 
ment of the belt on the pulley due to its elasticity and the dif- 
ference in tension between the tight and slack sides. Leather 
belting is extremely variable in its properties. The coefficient 
of friction for oak-tanned leather, hair side on a cast-iron turned 
pulley, ranges approximately as follows:* 

* Prof. Lanza, Trans. A. S. M. E., Vol. VII. See also paper by Wilfred Lewis, 
same volume and one by Prof. W. W. Bird in Vol. XXVI. Another good paper 
is that of F. W. Taylor, Vol. XV. 



290 



MACHINE DESIGN. 



in such a way that the necessary condition shall be fulfilled, and 
the belt will run properly. This gives what is known as a 
" twist" belt, and when the angle between the shaft becomes 
90 °, it is a " quarter-twist " belt. To make this clearer, see 
Fig. 174. Rotation is transmitted from A to B by an open belt, 
and it is required to turn the axis of B out of parallelism with 



1 E 



/ 
/ 

/ 
/ 

/ / 

/ 
/ 






y u 

Fig. 173 



v D' 




Fig. 174. 



that of A. The direction of rotation is as indicated by the 
arrows. Draw the line CD. If now the line CD is supposed 
to pass through the center of the belt at C and D, it may 
become an axis, and the pulley B and the part of the belt FC 
may be turned about it, while the pulley A and the part of the 
belt ED remain stationary. During this motion the center line 
of the part of the belt CF, which is the part that advances toward 
the pulley B when rotation occurs, is always in a plane perpen- 
dicular to the axis of the pulley B. The part ED, since it has 
not been moved, has also its center line in a plane perpendicular 
to the axis of A. Therefore the pulley B may be swung into 
any angular position about CD as an axis, and the condition 
of proper belt transmission will not be interfered with. 

169. If the axes intersect the motion can be transmitted 



BELTS, ROPES, BRAKES, AND CHAINS. 



291 



between them by belting only by the use of " guide" or "idler" 
pulleys. Let AB and CD, Fig. 175, be intersecting axes, and 
let it be required to transmit motion from one to the other by 
means of a belt running on the pulleys E and F. Draw center 
lines EK and FH through the pulleys. Draw the circle, G, 
of any convenient size, tangent to the lines EK and FH. In 
the axis of the circle, G, let a shaft be placed on which are two 
pulleys, their diameters being equal to that of the circle, G. 




Fig. 175. 



These will serve as guide-pulleys for the upper and lower sides 
of the belt, and by means of them the center lines of the advanc- 
ing parts of both sides of the belt will be kept in planes perpen- 
dicular to the axis of the pulley toward which they are advancing, 
the belts will run properly, and the motion will be transmitted, 
as required. 

The analogy between gearing and belting for the trans- 
mission of rotary motion has been mentioned in an earlier 
paragraph. Spur-gearing corresponds to an open or crossed 
belt transmitting motion between parallel shafts. Bevel-gears 
correspond to a belt running on guide-pulleys transmitting 
motion between intersecting shafts. Skew bevel and spiral gears 
correspond to a " twist" belt transmitting motion between shafts 
that are neither parallel nor intersecting. 



3°4 MACHINE DESIGN. 

This transforms into : 

Total friction work at both pulleys due to slippage in foot- 
pounds per minute 



H- 



+ 02). 



2 

Summing these three losses (a), (b), and (c) in foot-pounds 
per minute, subtracting them from PV and dividing the result 
by PV will give the efficiency of the drive. 

175. Problem. — A single-acting pump has a plunger 8 inches 
= 0.667 f° ot m diameter, whose stroke has a constant length of 
10 inches =0.833 f° ot - The number of strokes per minute is 50. 
The plunger is actuated by a crank, and the crank-shaft is 
connected by spur-gears to a pulley-shaft, the ratio of gears 
being such that the pulley-shaft runs 300 revolutions per minute. 
The pulley which receives the power from the line-shaft is 18 
inches in diameter. The pressure in the delivery-pipe is 106 lbs. 
per square inch. The line-shaft runs 150 revolutions per 
minute, and its axis is at a distance of 12 feet from the axis of 
the pulley-shaft. 

Since the line-shaft runs half as fast as the pulley-shaft, the 

diameter of the pulley on the line-shaft must be twice as great 

as that on the pulley-shaft, or 36 inches. The work to be done 

per minute, neglecting the friction in the machine, is equal to 

the number of pounds of water pumped per minute multiplied 

by the head in feet against which it is pumped. The number 

of cubic feet of water per minute, neglecting "slip," equals the 

displacement of the plunger in cubic feet multiplied by the 

0.667 2 Xtt 

number of strokes per minute = X 0.833X50 = 14.5 5, 

4 

and therefore the number of pounds of water pumped per minute 
= 14.55X62.4 = 908. One foot vertical height or "head" of 
water corresponds to a pressure of 0.433 lb. per square inch, 
and therefore 100 lbs. per square inch corresponds to a "head" 



BELTS, ROPES, BRAKES, AND CHAINS. 



293 



made in the variables R and r, so long as their sum is constant, 
will not affect the value of the equation, and hence the belt 
length will be constant. It will now be easy to design cone 
pulleys for a crossed belt. Suppose a pair of steps given to 
transmit a certain velocity ratio. It is required to find a pair 
of steps that will transmit some other velocity ratio, the length 
of belt being the same in both cases. Let R and r = radii of the 
given steps ; R' and r' = radii of the required steps ; R + r = 
R'+r'=a; the velocity ratio of R' to r' =b. There are two 




Fig. 177. 

equations between R' and r', R f -r-r' =b, and R' + r'=a. Com- 
bining and solving, it is found that r' = a^-(i+6), and R! =a — r'. 
For an open belt the formula for length, using same symbols 
as for crossed belt, is 



L = 2 Vd 2 -(R-r) 2 + 7r(R + r) 

+ 2(R — r) (arc whose sine is (it — r) -f- d). 

If R and r are changed as before (i.e., R + r = R' + r' =a 
constant), the term R — r would of course not be constant, and 
two of the terms of the equation would vary in value; therefore 
the length of the belt would vary. The determination of cone 
steps for open belts therefore becomes a more difficult matter, 
and approximate methods are almost invariably used. 

172. Graphical Method for Cone-pulley Design. — The fol- 
lowing graphical approximate method is due to Mr. C. A. Smith, 
and is given, with full discussion of the subject, in "Transactions 



294 



MACHINE DESIGN. 



of the American Society of Mechanical Engineers," Vol. X, 
p. 269. Suppose first that the diameters of a pair of cone steps 
that transmit a certain velocity ratio are given, and that the 
diameters of another pair that shall serve to transmit some 
other velocity ratio are required. The distance between centers 
of axes is given. (See Fig. 178.) Locate the pulley centers O 



K B 




Fig. 178. 

and O r at the given distance apart; about these centers draw 
circles whose diameters equal the diameters of the given pair of 
steps; draw a straight line GH tangent to these circles; at /, 
the middle point of the line of centers, erect a perpendicular, 
and lay off a distance JK equal to the distance between centers, 
C, multiplied by the experimentally determined constant 0.314; 
about the point K so determined, draw a circular arc AB tan- 
gent to the line GH. Any line drawn tangent to this arc will be 
the common tangent to a pair of cone steps giving the same 
belt length as that of the given pair. For example, suppose 
that OD is the radius of one step of the required pair; about O, 
with a radius equal to OD, draw a circle ; tangent to. this circle 
and the arc AB draw a straight line DE; about O' and tangent 
to DE draw a circle \ its diameter will equal that of the required 
step. 

But suppose that, instead of having one step of the required 
pair given, to find the other corresponding as above, a pair of 



BELTS, ROPES, BRAKES, AND CHAINS. 295; 

steps are required that shall transmit a certain velocity ratio r 

= r, with the same length of belt as the given pair. Suppose 

OD and O'E to represent the unknown steps. The given velocity 

OD 
ratio equals r. Also, T^'KTB- But from similar triangles 

FO 
OD^O'E=FO+FO'. Therefore r=^^; but FO=C + x y 

rU 

C + x C 

and FO' =x. Therefore r= — , and x = . Hence with 

x r—i 

r and C given, the distance x may be found, and a point F located^ 

such that if from F a line be drawn tangent to AB, the cone 

steps drawn tangent to it will give the velocity ratio, r, and a 

belt length equal to that of any pair of cones determined by a 

tangent to AB. The point F often falls at an inconvenient 

distance. The radii of the required steps may then be found 

as follows: Place a straight-edge tangent to the arc AB and 

measure the perpendicular distances from it to O and O'. 

The straight-edge may be shifted until these distances bear 

the required relation to each other. In this case it is well to check 

the accuracy of the construction by computing the resultant 

length of belt with each pair of steps. 

173. Design of Belts.— Fig. 179 represents two pulleys 

connected by a belt. When no moment is applied tending to 

produce rotation this tension in the two sides of the belt is 

practically equal. Let T represent this tension. If now 

an increasing moment, represented by Rl, be applied to the 

driver, its effect is to increase the tension in the lower side of 

the belt and to decrease the tension in the upper side. With 

the increase of Rl this difference of tension increases till it is 

equal to P, the force with which rotation is resisted at the surface 

of the pulley. Then rotation begins * and continues as long as 

* While the moving parts are being brought up to speed the difference of ten- 
sion must equal P plus force necessary to produce the acceleration. 



296 



MACHINE DESIGN, 



this equality continues; 'i.e., as long as T 1 -T 2 = P, in which 
T 1 = tension in the driving side and T 2 = tension in the slack 
side. (See Fig. 180.) The tension in the driving side is increased 
at the expense of that in the slack side, but the sum does not 
remain a constant. Analysis of experimental data * shows that 
a close approximation is given by the simple equation 

Vt\+Vt 2 =2Vto. 

T 
To find the value of — . The increase in tension from the 
J- 2 

slack side to the driving side is possible because of the frictional 

resistance between the belt and pulley surface. Consider any 

Fig. 179. 




Fig. 



element of the belt, ds, Fig. 181. It is in equilibrium under 
the action of the following forces: 

T, the value of the varying tension at one end of ds; 

T + dT, the value of the varying tension at the other end of ds ; 

* Wilfred Lewis in Trans. A. S. M, E„ Vol, VII, 



BELTS, ROPES, BRAKES, AND CHAINS. 



297 



cds, the centrifugal force; 

pds, the pressure between the face of the pulley and ds\ 

dF = /j.pds, the friction between the element of belt and pulley 
face. 

These correspond to any cross- sectional area, A, square 
inches. 




Tt is more convenient to develop the equations if a cross- 
sectional area of one square inch is considered. For such a 

/ T , \ /T" 1 I /7TA 

section the forces acting become ^ = (-tL t + dt = l ), cds, 

pds, and /ipds. 



298 MACHINE DESIGN. 

T 

Let h — — ■= allowable tension in tight side of belt, pounds 

per square inch. 240 lbs. is recommended as most 

economical value. 

T 2 
t2 = — - = tension in slack side, pounds per square inch; 

P 
ti —t 2 = — = effective pull, pounds per square inch; 

v = velocity of belt in feet per second; 
V= velocity of belt in feet per minute; 
w = weight of belt, pounds per cubic inch; 
c = centrifugal force per cubic inch of belt at velocity 

(l2WV 2 ^ 

V gr 
r = radius of pulley in inches ; 

r 
R = radius of pulley in feet = — ; 

12 

a = arc of contact in degrees ; 

= arc of contact in radians = 0.01 75a; 

T 
to=—7 = initial tension, both sides, pounds per square inch; 
A. 

T c 
t c =~r = centrifugal tension per square inch of cross- 

(l2WV 2 \ 

section = ) ; 

\ g V 

f = pressure per linear inch between pulley and belt; 
/£ = coefficient of friction. 

Summing the vertical components: 

. dO f j:a . \dd, 

pas + cds = t sin \-\t-\- at) sin — . 

2 2 

dd dd 

dd is so small that sin — may be considered equal to — . 

2 2 



BELTS, ROPES, BRAKES, AND CHAINS. 299 

Also dt and dO being very small compared to the other quan- 
tities, any terms containing their product may be dropped. 
Therefore 

pds+cds = tdd, 
but 



and 



ds = rdd. 



\ cds = -dd = t c dd, 



.-. pds=(t-t c )dd. 
Summing the moments about O : 

t + dt = t+/jLpds, 
dt = [tpds 
= //(/ -t c )dd, 

P 1 dt r e ln 

Ju t-t c Vo 

1 tl -<■ a 

loge; T=^> 

t2 ~t c 

common log 1—1=. ^^pd. 

h ~t c 

The following equations are now established : 



fc-fc-ji w 

vr l +vt 2 = 2 vt , . (2) 

log; 1 — -f = . 4343/^- ..... (3) 



300 MACHINE DESIGN. 

For speeds below 1800 ft. per minute equation (3) may be 
written as 

h 

log - = .4343^ (30 

h 

To apply these equations it is simplest to decide upon a max- 
imum value of t\. This varies with the quality of the belt, the 
nature of the splice, etc., and may be taken as high as 300 lbs. 
per square inch; but when the economic life of the belt is con- 
sidered, 200 to 240 lbs. per square inch is better. For a new 
belt take 240 lbs. 

It is evident that the foregoing equations may also be written 
in the following forms : 

T 1 -T 2 =P, (ia) 

VT 1 -Vt 2 =2\ / t , ( 2 a) 

log * ' = .4343/^ (3 a ) 

1 2 -1 c 

and for low velocities 

T 

log -=^=.4343/^ (3<*') 

J- 2 

The weight of ordinary oak-tanned leather belting per cubic 
inch may be taken, w= 0.035. 

Table XIX. 

\T 1 r , T-2WV 2 

Values of tc = . 



V 


1800 


2400 


3000 


3600 


4200 


4800 


5400 


6000 


6600 


7200 


7800 


V 


30 


40 


So 


60 


70 


80 


90 


100 


no 


120 


130 


tc 


11. 7 


20.8 


32.6 


47.0 


64.0 


83.2 


106 


130. 5 


IS« 


188 


221 



From Table XIX it becomes evident that the centrifugal 
tension, which diminishes the effective tension- producing pressure 
between belt and pulley upon which the frictional driving power 



BELTS, ROPES, BRAKES, AND CHAINS. 301 

depends, increases rapidly with the velocity and if t\ = 240 lbs. 
per square inch there will be no effective pressure at a speed of 
about 8000 ft. per minute. In other words the belt, if put on 
with the proper value of / corresponding to ^ = 240, can trans- 
mit no power at this speed, because the centrifugal force is so 
great that no pressure exists between the belt and the face of 
the pulley, and hence there is no friction. 

The necessary value of P is a given condition in any problem. 
If the power to be transmitted by the belt is given in HP and 
the velocity of the belt, V, in feet per minute is known, 

HPX 33,000 
V 

The most economical speed at which to use a leather belt is 
about 4500 ft. per minute. In general P is determined by 
dividing the foot-pounds of work per minute to be transmitted, 
by the belt speed (or pulley rim velocity) in feet per minute. 

The value of is determined for the pulley of smallest arc 
of contact from the diameters of the pulleys and the distance 
between their centers. (See sec. 174.) 

The value of //, the coefficient of friction, varies with the 
kind of belting, the material and character of surface of pulley, 
the condition of the belt as regards dressing, the side of the 
leather used, and particularly with the rate of slip of the belt 
on the pulley. This slip is a compound of two factors, actual 
slippage and belt creep, the latter being the unavoidable move- 
ment of the belt on the pulley due to its elasticity and the dif- 
ference in tension between the tight and slack sides. Leather 
belting is extremely variable in its properties. The coefficient 
of friction for oak-tanned leather, hair side on a cast-iron turned 
pulley, ranges approximately as follows:* 

* Prof. Lanza, Trans. A. S. M. E., Vol. VII. See also paper by Wilfred Lewis, 
same volume and one by Prof. W. W. Bird in Vol. XXVI. Another good paper 
is that of F. W. Taylor, Vol. XV. 



302 



MACHINE DESIGN. 



M 


Av. vel. of slip on one 
pulley, ft. minute =Vs 


.12 tO .17 


O 


.24 
.28 


2.1 
2.6 


•31 


6.9 


■33 


15 


.82 


2IO 



This corresponds roughly to 
/*=i.oo 



48 



6i + F s 

Prof. Lanza recommends a uniform value of ^=.27. This 
corresponds to a uniform slip of between two and three feet 
at all speeds. 

In his valuable paper in Trans. A. S. M. E., Vol. XXXI, 
Carl Barth writes this formula, based upon his own experiments 
and those of Prof. Bird, 

and he also recommends, 



4+v; 



^=.54 



140 



500 +v 
Equating these two expressions for /i, 



V s = 



160+0.88F 



85+0.03F 

These formulae are open to criticism, but may be accepted 

tentatively until further data are made available. 

The total slip on two pulleys = 2 V s , .'. per cent of total 

200 V s 
shp= — . 

174. Efficiency of Belt Drive. — An approximate estimate 
of the efficiency of a belt drive can be made. The subscripts 
1 and 2 are used for driver and follower, respectively. There 



BELTS, ROPES, BRAKES, AND CHAINS. 303 

are three losses to be considered: (a) journal friction, (b) work 
of bending the belt, (c) slip and creep friction. Each of these 
losses takes place at each pulley. 

(a) The journal friction work per minute at both bearings, 
in foot-pounds 

where //' = coefficient of journal friction appropriate to 

conditions (see Chaps. XII and XIII) ; 
d\ and d 2 = journal diameters, inches; 
iVi and N 2 = revs. per minute; 
Ti, T 2 , and T c , total belt tensions as above, pounds. 

(b) For the work lost in bending the belt the following formula, 
based upon Eytelwein's for ropes, may be used in default of 
one based upon specific investigation. 

Work in foot-pounds per minute bending belt 



= .04SPV 



Vi rj 



P = belt pull=7\ -T 2 , in lbs.; 
V = belt velocity, feet per minute ; 
h = belt thickness, inches; 
r x and r 2 = pulley radii, inches. 

(c) The friction loss due to belt slippage at each pulley in 
foot-pounds per minute 

= HprdV s . 

/i = coefficient of friction of belt on pulley at the selected 

rate of slip; 
p = pressure between belt and pulley face per linear inch, 

pounds; 
r = radius of pulley, inches; 
0=--2ltc of contact, radians; 
F 5 = slip at each pulley, feet per minute. 



3°4 MACHINE DESIGN. 

This transforms into : 

Total friction work at both pulleys due to slippage in foot- 
pounds per minute 



=MV S {- 



(01+02). 



Summing these three losses (a), (b), and (c) in foot-pounds 
per minute, subtracting them from PV and dividing the result 
by PV will give the efficiency of the drive. 

175. Problem. — A single-acting pump has a plunger 8 inches 
= 0.667 foot in diameter, whose stroke has a constant length of 
10 inches =0.833 f° ot - The number of strokes per minute is 50. 
The plunger is actuated by a crank, and the crank-shaft is 
connected by spur-gears to a pulley-shaft, the ratio of gears 
being such that the pulley-shaft runs 300 revolutions per minute. 
The pulley which receives the power from the line- shaft is 18 
inches in diameter. The pressure in the delivery-pipe is 100 lbs. 
per square inch. The line-shaft runs 150 revolutions per 
minute, and its axis is at a distance of 12 feet from the axis of 
the pulley-shaft. 

Since the line-shaft runs half as fast as the pulley- shaft, the 

diameter of the pulley on the line-shaft must be twice as great 

as that on the pulley-shaft, or 36 inches. The work to be done 

per minute, neglecting the friction in the machine, is equal to 

the number of pounds of water pumped per minute multiplied 

by the head in feet against which it is pumped. The number 

of cubic feet of water per minute, neglecting "slip," equals the 

displacement of the plunger in cubic feet multiplied by the 

0.66 fXn 

number of strokes per minute = Xo.833X5o = i4.55, 

4 

and therefore the number of pounds of water pumped per minute 
= 14.55X62.4 = 908. One foot vertical height or "head" of 
water corresponds to a pressure of 0.433 lb. P er square inch, 
and therefore 100 lbs. per square inch corresponds to a "head" 



BELTS, ROPES, BRAKES, AhID CHAINS. 305 

of 100-^0.433 = 231 feet. The work done per minute in pump- 
ing the water therefore is equal to 908 lbs. X 231 ftet =209,748 
ft. -lbs. The velocity of the rim of the belt-pulley is equal to 
300x1.5 X~= 1414 feet per minute.* Therefore the force P - T\ 
— 7^2 = 209,748 ft. -lbs. per minute -M414 feet per minute = 
148 lbs. 



Fig. 182. 

j, „ n // 

To find a (see Fig. 182) sin /?- 2 = - = 0.0625. 

/ 144 

Therefore ,3 = 3° 35'; a = i8o o -20 = i8o°-7° 10' = 172° 50'; a 

in 7z measure = 172! X 0.0175 =3.025 = 0. 
T 

log y~ =0.4343 xnxd =0.4343x0.3x3.025 =0.3941. 

.-. ^ = 2.48; p = r 1 -r 2 = i 4 8. 
j- 2 

Combining these equations T\ is found to be equal to 248 lbs., 
the maximum stress in the belt. 

7\ 248 
The cross-sectional area of belt should be equal to — = — 

t\ 240 

= 1.03 square inch. 

Single-thickness belting varies from 0.2 to 0.25 of an inch in 

thickness, hence the width called for by our problem would be 

1.03 . , 

= 5 inches, say. 

0.2 

176. Problem. — A sixty-horse-power dynamo is to run 1500 

revolutions per minute and has a 15-inch pulley on its shaft. 

* At this speed the simple form of the belt formula may be used. 



306 machine design. 

Power is supplied by a line-shaft running 150 revolutions per 
minute. A suitable belt connection is to be designed. 

The ratio of angular velocities of dynamo-shaft to line-shaft 
is 10 to 1; hence the diameter of the pulley on the line-shaft 
would have to be ten times as great as that of the one on the 
dynamo, =12.5 feet, if the connections were direct. This is 
inadmissible, and therefore the increase in speed must be ob- 
tained by means of an intermediate or counter shaft. Suppose 
that the diameter of the largest pulley that can be used on the 
counter-shaft =48 inches. Then the necessary speed of the 

counter-shaft = 1 500 X -5 =470, nearly. The ratio of diameters 

48 

of the required pulleys for connecting the line-shaft and the 

470 
counter-shaft =- — = 3.13- Suppose that a 60-inch pulley can 

be used on the line-shaft, ( then the diameter of the required 

60 
pulley for the counter-shaft will = =19 inches, nearly. Con- 

sider first the belt to connect the dynamo to the counter-shaft. 
The work = 60X33,000 = 1,980,000 ft. -lbs. per minute; the rim 

7TI5 

of the dynamo-pulley moves X 1500 = 5890 feet per minute. 

1980000 

Therefore T\ — T 2 = — o = n6 lbs. The axis of the counter- 

5890 ^ 

shaft is 10 feet from the axis of the dynamo, and, as before, 

. Q r 1 -r 2 24-7.5 
sin0« — -__ -0.1375. 

Therefore fi=f 54'. 

a = i8o°-2/?=i64° 12', 
#=i64°.2Xo.oi75 = 2.874. 

The nearest value of V, in Table XIX, to 5890 is 6000, and 
the corresponding value of t c is 130.5. 
More accurately 

4=126, 



BELTS, ROPES, BRAKES, AND CHAINS. 307 

140 

500 + 5890 

240-126 
log -=0.4343X0.518X2.874 

f 2 -120 

= 0.6442. 

n4 = 4-3 8 7fe-i26), 

'2 = 152, 

/1 -/ 2 = 240 -152 = 88, 

^ 336 

^4 = = — — = 3.82 square inches. 

h -k 88 ° H 

A double belt is about f inch thick. Our problem, then, 

calls for a double belt 1X3.82 = 10 inches, say, wide. 

177. Variation of Driving Capacity. — From equation ($a'), 

T 
sec. 173, it follows that the ratio of tensions, — , when the belt 

slips at a certain allowable rate (i.e., when ft is constant), de- 
pends only upon a. The velocity of the belt also remains 
constant. This ratio, therefore, is independent of the initial 

T 

tension, T ; hence "taking up" a belt does not change — -. 

T2 

The difference of tension, 7\ -T 2 = P, is, however, dependent 

on T . Because p, the normal pressure between belt and 

pulley, varies directly as T , then, since dF ' = upds = dT \ it 

follows that dT varies with T ; and hence 



^dT=T Y -T 2 = P 



varies with T . This is equivalent to saying that " taking 
up " a belt increases its driving capacity. 

This result is modified because another variable enters the 
problem. If T is changed, the amount of slipping changes, 
and the coefficient of friction varies directly with the amount 
of slipping. Therefore an increase of T would increase p 



308 MACHINE DESIGN. 

and decrease // in the expression /ipds=dT, and the converse 
is also true. This is probably of no practical importance. 

The value of P may also be increased by increasing either /*, 
the coefficient of friction, or 6, the arc of contact, since increase 

T 

of either increases the ratio — , and therefore increases 

J- 2 

T 1 -T 2 =P. 

Increasing To decreases the life of the belt. It also in- 
creases the pressure on the bearings in which the pulley-shaft 
runs, and therefore increases frictional resistance; hence a 
greater amount of the energy supplied is converted into heat and 
lost to any useful purpose. But if T is kept constant, and /* 
or is increased, the driving power is increased without noticeable 
change of pressure in the bearings, since V7\ + Vt 2 remains 
constant. When possible, therefore, it is preferable to increase 
P by increase of pt or 6, rather than by increase of T . 

Application of belt-dressing may serve sometimes to in- 
crease fi. 

If, as in Fig. 179, the arrangement is such that the upper 
side of the belt is the slack side, the "sag " of the belt tends to 

T 

increase the arc of contact, and therefore to increase ~^r. If 

J- 2 

the lower side is the slack side, the belt sags away from the pulleys 

T 

and d and tpt are decreased. 
J- 2 




Fig 183. 
An idler-pulley, C, may be used, as in Fig. 183. It is pressed 
against the belt by some means. Its purpose may be to increase 

P by increasing the tension, T =i I £j 2 . In this case 






BELTS, ROPES, CHAINS, AND BRAKES. 



3°9 



friction in the bearings is increased, and this method should be 
avoided. Or it may be used on a slack belt to increase the 

T 

angle of contact, «, the ratio — , and therefore P, the driving 

^2 



force. In this case the value of T ( 



_/ vr 1 +vr 2 \ 



may be 



made as small a value as is consistent with driving, and hence 
the journal friction may be small. 

Tighteners are sometimes used with slack belts for dis- 
engaging gear, the driving-pulley being vertically below the 
follower. 

In the use of any device to increase fi and a, it should be 
remembered that T\ is thereby increased, and may become 
greater than the value for which the belt was designed. This 
may result in injury to the belt. 

In Fig. 184, the smaller pulley, A, is above the larger one, 
B. A has a smaller arc of contact, and hence the belt would 




Fig. 184. Fig. 185. 

slip upon it sooner than upon B. The weight of the belt, how- 
ever, tends to increase the pressure between the belt and A, 
and to decrease the pressure between the belt and B. The 
driving capacity of A is thereby increased, while that of B is 
diminished; or, in other words, the weight of the belt tends to 
equalize the inequality of driving power. If the larger pulley 
had been above, there would have been a tendency for the 
belt weight to increase the inequality of driving capacity of the 



310 MACHINE DESIGN. 

pulleys. The conclusion from this, as to arrangement of pulleys, 
is obvious. 

178. Proper Size of Pulleys. — A belt resists a force which 
tends to bend it. Work must be done, therefore, in bending a 
belt around a pulley. The more it is bent the more work is 
required and the more rapidly the belt is worn out. Suppose 
AB, Fig. 185, to represent a belt which moves from A toward 
B. If it runs upon C it must be bent more than if it runs 
upon D. The work done in bending the belt is converted into 
useless heat by the friction between the belt fibers. It is desirable, 
therefore, to do as little bending as possible. This is one reason 
why large pulleys in general are more efficient than little ones. 
The resistance to bending increases with the thickness of the 
belt, and hence double belts should not be used on small pulleys 
if it can be avoided. 

Double belts may be used on pulleys 12" and over. 
Triple " " " " " " 20" " 

Quadruple " " " " " " 30" " " 

179. Distance Desirable between Shafts. — In the design of 
belting care should be taken not to make the distance between 
the shafts carrying the pulleys too small, especially if there is the 
possibility of sudden changes of load. Belts have some elasticity, 
and the total yielding under any given stress is proportional to 
the length, the area of cross-section being the same. There- 
fore a long belt becomes a yielding part, or spring, and its yielding 
may reduce the stress due to a suddenly applied load to a safe 
value ; whereas in the case of a short belt, with other conditions 
exactly the same, the stress due to much less yielding might be 
sufficient to rupture or weaken the joint. 

180. Rope-drives. — The formulae which have been derived 
for belts also apply to rope-drives. For good durability the 
allowable tension in a rope-drive should be about 20od 2 lbs. 
where d is the diameter of the rope in inches. Experiments 



BELTS, ROPES, CHAINS, AND BRAKES. 



3" 



vary greatly in the value of the coefficient of friction for a well- 
lubricated rope on a flat-surfaced smooth metal pulley. It may 
be taken equal to 0.12.* But ropes are not commonly used 
on flat pulleys; instead of this they run in grooves on the faces 
of sheave wheels, and substitution must be made for // in the 



formula (3), not 0.12 but o.i2Xcosec 



angle of groove 



The following table gives the values of /i for different angles 
of grooves. 





Table XX 












Angle of groove in degrees . . . 


3° 
0.46 


35 
0.40 


40 
0-35 


45 
0.31 


5° 
0.28 


55 
0.26 


60 
0.24 







Fibrous ropes for power transmission purposes are made 
chiefly of cotton or manilla fiber. The former is softer, more 
flexible, and elastic, the latter is cheaper and stronger. The 
former weighs about 10 per cent less than the latter. The fol- 
lowing tables are computed for manilla rope. 

Taking the weight of rope per linear inch = o.o268d 2 lbs., 
and the allowable tension, T 1 = 2ood 2 lbs., and solving for T e 
at various speeds, gives the following results : 

Table XXL 
(rc = 12^=0.01^.) 



1000 2000 
2.78,11.1 



2500 
17-4 



3000 
25.0 



35oo 
340 



4000 
44-4 



4500 
56.2 



5000 
69-5 



55oo 
84.0 



6000 
100 



6500 
118 



7000 
136 



7500 
156 



8000 
178 



8500 
200 



V is the velocity of the rope in feet per minute. 
For convenience the following table is given showing the cor- 
responding values of angles in degrees and circular measure : 



* "Rope-driving," by J. J. Flather. New York: Wiley & Sons. 



312 



MACHINE DESIGN. 



Table XXII. 



I°5 
1.83 


120 
2.09 


135 
2.35 


150 

2.62 


165 

2.88 


180 
3.14 


195 
3-43 


210 
3.66 



240 

4.19 



It will be remembered that #=0.01 75a. 

The diameter of the sheave wheel is properly calculated from 
the point of tangency of the rope to the groove (A- A, Fig. 186) 
and not from the middle of the rope O. The diameter of the 
sheave wheel should not be too small or the rope will wear out 
very rapidly. The following table gives the minimum values 
D being the diameter of the wheel and d that of the rope : 

Table XXIII 



& 


3rr 

4 


l" 


i£" 




if" 


2 " 


D 


24 


36 


48 


6O 


72 


84 



Table XXIII is for general purposes. Specifically the formula 
D = d 1 - 7 X'VV + i2 inches may be used. 

Two systems of rope-driving are in use, the English and 
the American. In the former a number of ropes are used side 
by side. In the latter a single continuous rope is used with 
a guide and tightener. As long as all the grooves in each 
sheave are alike, each rope will tend to carry its proportionate 
share of the load in the English system, provided all the ropes 
had the same original tension, and have stretched the same 
amount. 

Next to the angle of groove the most important item is to 
have the grooves as smoothly surfaced as possible. 

Fig. 1S6A shows one manufacturer's standard groove pro- 
portions for fibrous ropes. 

The diameters of multiple-grooved sheaves must be accurately 
alike. 

For the American system the grooves In tne larger sheave 



BELTS, ROPES, BRAKES, AND CHAINS. 



313 



should have a greater angle than those in the smaller sheave, or the 
load will be unequally divided among the various wraps of the rope.* 






\ 



-a o 







O^ 



U) 







CO +3 

to 

oco 



9 

-2 « 

^£ 
c 
o 

en 



vo 
OO 



u 



The tension in each wrap of the rope will be the same, when 
running, if the friction on each sheave wheel is the same. The 
friction on each pulley will be the same if the products of the 

arcs of contact by the respective coefficients are equal. 

y 

Let [i = coefficient of larger sheave = 0.1 2 cosec — ; 

2 
Y = angle of groove of larger sheave; 

a = arc of contact of larger sheave ; 

* See further Proc. Am. Soc. C. E., Vol. XXIII. Mr. Spencer Miller on 
Rope-driving. 



314 



MACHINE DESIGN. 



uf = coefficient of smaller sheave =0.12 cosec — : 

r 2 

f = angle of groove of smaller sheave; 
a' =arc of contact of smaller sheave. 

Then pa should = //a\ 

r f a f 

.'. cosec — = cosec — X— . 
2 2 a 

The following table gives the proper values for equal adhesion 

Table XXIV. — Angle of Groove for Equal Adhesion. 



Arc of contact on small pulley a' 
Arc of contact on large pulley a 
Angle of groove in large pulley when 

groove in small pulley =35° 

Angle of groove in large pulley when 

groove in small pulley = 40 

Angle of groove in large pulley when 

groove in small pulley = 45° 



0.9 


0.8 


o.75 


0.7 


0.65 


40 


44° 


47° 


5i° 


55° 


45° 


5o° 


54° 


5 8° 


6 4 ° 


5o° 


55° 


6o° 


66° 


72° 



0.6 

6o° 
70° 
8o° 



The angle of groove on the smaller sheave wheel is generally 
made 45 . Assuming this an angle of contact of 165 , and an 
allowable stress = 2ood 2 , the following table has been com- 
puted for the horse-power transmitted by each wrap of the rope : 

Table XXV. — Horse Power Transmitted by Single Rope. 



Velocity 
















of rope in 
















ft. per 






Diameter of rope in 


inches. 






mm. 
















V. 


! 


1 

4 


1 


X * 


15 


if 


2 


1000 


1.38 


1.98 


3-52 


5-50 


7.91 


II . 20 


14.08 


2000 


2 


64 


3.80 


6-75 


10.55 


15.20 


21.50 


27.00 


2500 


3 


18 


4-58 


8.15 


12.70 


18.35 


26.00 


32.60 


3000 


3 


67 


5-28 


9.40 


14.70 


21.15 


30.00 


37.60 


3500 


4 


07 


5-85 


IO.40 


16.25 


23.40 


33- 20 


41.60 


4000 


4 


36 


6.28 


11. 15 


17.40 


25.10 


35- 60 


44.60 


4500 


4 


54 


6-54 


11 .60 


18.12 


26. 10 


37.00 


46.40 


5000 


4 


57 


6-59 


11.70 


18.30 


26.35 


37-3o 


46.80 


5 Soo 


4 


45 


6.42 


11 .40 


17.80 


25-65 


36.40 


45- 60 


6000 


4 


2 


6.05 


I0.75 


16.80 


24. 20 


34 30 


43.00 


6500 


3 


73 


5-37 


9-55 


14-93 


21.50 


30.40 


38.20 


7000 


3 


12 


4-5o 


8.00 


12.50 


18.00 


25-50 


32.00 


7500 


2 


3i 


3-32 


5 90 


9.21 


I3-30 


18.80 


23.60 



BELTS, ROPES, BRAKES, AND CHAINS. 315 

For durability a few turns of a larger rope are preferable to 
more turns of a smaller rope. 

The most economical speed, taking first cost and relative wear 
into consideration, is about 4500 feet per minute. 

In any given case, since T\ = 2ood 2 and T c = o.oiv 2 d 2 , T 2 
can be computed by writing for it xd 2 . Substituting these values 
in equation (3), 

200^ 2 -o.oiv 2 d 2 
l0g **-o.oi^ "°-4343^. 

The term d 2 divides out, x can be solved for, and the value 
of T 2 determined from T 2 = xd 2 . 

The initial tension, T , is determined from the equation 

The length and deflections of the rope are important points. 
The curve the rope takes between supports is the catenary and 
the following equations . are approximately correct for horizontal 
drives :* 

C 2 w 

A — deflection at middle of span in feet; 

C=span, in feet; 

w = weight of rope per linear foot; 

* For inclined drives, etc., see Reuleaux's Constructor or Flather's Rope Driving. 
Trade catalogs of the C. W. Hunt Co., the Plymouth Cordage Co., and others, 
give valuable data on groove forms, arrangement of installations, etc. 



316 MACHINE DESIGN. 

T= tension in pounds (T , T Xi or T 2 for A , A lt or A 2 ) ', 
L = length of catenary in feet. For original length use Tq 
and A . The entire rope length for one wrap = 
2L Q + R l d 1 +R 2 d 2 ; 
Ri, R 2 = pulley radii, in feet; 
61, 2 = corresponding arcs of contact, radians. 

181. Efficiency of Rope Drives. — The claims made for rope 
driving embrace: suitability to transmitting large amounts of 
power, quiet running, can be carried in any direction, can be 
subdivided most readily, does not require accurate alignment of 
sheaves, freedom from electrical disturbance, reasonably weather 
proof, economy in first cost and in maintenance. E. H. Ahara, in 
Trans. A. S. M. E., Vol. XXXV, reports on a series of about 700 
tests, extending over a continuous period of five months. His 
results show higher efficiencies for the American system than 
for the English and higher efficiencies for the open drive than 
the " up and over," with either system. He used extreme belt 
tensions, in some cases T 2 = ^6od 2 , therefore transmitting as much 
as four times the power suggested as economical in Table XXV. 
This was done without loss in efficiency, whatever may have 
been the result on the ultimate life of the rope. Data collected 
by J. J. Flather indicate that ropes in which 7\ greatly exceeds 
2ood 2 wear out with serious rapidity. The Ahara tests show 
American open drives under best conditions averaging about 
93 per cent, English open drives 87 per cent, American " up 
and over " 80 per cent, English " up and over," 75 per cent. 
They seem to show " that the efficiency in rope driving is con- 
siderably greater at the lower speeds than at the higher ones, 
the dropping off being especially noticeable above 4500 ft. per 
minute of rope speed. They also show that the efficiency of a 
rope drive is not materially affected by distances between centers 
up to 150 ft., that the drop of efficiency at 50 per cent load is 
comparatively small over that of full load, and that, if proper 



BELTS, ROPES, BRAKES, AND CHAINS. 317 

care is exercised to have all grooves perfect in pitch diameter, 
many as well as few ropes can be run on a drive with good 
efficiency." 

182. Problem. — An engine, running at 100 revs, per minute 
and delivering 275 H.P. is to drive a main-shaft 60 ft. distant 
at 300 revs, per minute. Conditions permit the use of sheaves of 
most economical size. 

Select 7 = 4500 ft. per minute. 

V 

Engine sheave diameter = =14.201 ft. 

5 ttXioo y 

14.201 
Shaft sheave diameter = = 4.764 ft. 

3 

Allowable rope diameter, from 

J D = ^ 1J v / F + i2ins., 
where 

D = smaller sheave diameter in inches, ^ = 1.5 inches. 

Angle of contact on smaller pulley = 

« « . 7-146 -2.382 
180 -2 sin -1 ' £ — — = i7i° = 3 radians. 
60 

Angle of contact, larger pulley = 3.283 radians. 
T e from Table XXI = 56.2^. ^=0.31. 

200-56.2 
" g x- S 6.2 =°-4343Xo.3iX3. 

#=113. T 2 =nsd 2 . 

T 1 -T 2 ^= 20od 2 - 1 i 3 d 2 = Sjd 2 . 
= 196 lbs. 

tt -r. 4500XI96 

H.P. per wrap= — — = 26.7. 

33,000 



3 1 8 MACHINE DESIGN, 

275 
Number of wraps = — — = 10.5, say 10 wraps. 

T =i K x - 2 J = i 3 8*= 3 iolbs., 

6o 2 X 0.322* 

J ° = 8X138* =I - 05ft " 
z0=6o+ 8 ^ =6o . 5ft . 

3X60 ° 

Length of one wrap (under tension T ) = 

2X60.5 + 7.146X3.283 + 2.382X3 = 151.6 ft. 

Length of ten wraps = i5i6 ft. 

Add allowance for splice and tension carriage. 

6o 2 X 0.322* 
Jl= SXsocxF =- 725ft -' 

8XII3* 

183. Wire Rope Transmission.* — For many years wire rope 
has been used satisfactorily for power transmission. It is not 
particularly applicable to short spans (where they are under 
60 ft. it is not possible to splice the rope with such a degree of 
nicety as to give the desired tensions and deflections, and some 
mechanical adjustment becomes necessary) but it is applicable 
to long spans. At Lockport, N. Y., a clear span of 1700 ft., 
without intervening support, has been used. The length of clear 
span is determined by the allowable deflections. When the 
distance exceeds the limit for a clear span, supporting sheaves 

* See Reuleaux's Constructor. 



BELTS, ROPES, CHAINS, AND BRAKES. 319 

with unfilled grooves (idlers) are used. In very long trans- 
missions it is impracticable to run the rope at high velocities. 
The force factor must therefore be made greater, and this is 
done by increasing the angle of contact by lapping the rope 
several times about a pair of grooved drums at each end. The 
sheaves used should be filled with hard wood, tarred oakum, 
or segments of leather and rubber soaked in tar and packed 
alternately in the groove. The sheaves should be accurately 
balanced. The rope rests free on the packing and does not 
wedge in the groove as with fibrous rope transmission. 

The same general formulae apply as developed for fibrous 
ropes. 

Under ordinary conditions, six-strand ropes of seven wires 
to the strand, laid about a hemp core, are best adapted to the 
transmission of power, but conditions often occur where twelve- 
wire or nineteen-wire rope is to be preferred. 

The weight of cast-steel rope per linear foot is approximately, 

~d 2 . ... 

w= — . Its ultimate tensile strength is approximately 64,000^. 

For plough steel this is 76,000^ to 90,000^. For Swedish iron, 
30,oooJ 2 . 

The stress induced by bending the rope around the sheave 
is computed from the formula (Bach), 



Ed 



/ 6 = flexural stress, pounds per square inch; 

E = modulus of elasticity, 29,000,000 for steel; 

D = diameter of sheave, inches; 

d = diameter of individual wires, inches 
= I diam. of rope for 7-wire rope 
=t6 diam. of rope for 19-wire rope. 



3 2 ° 



MACHINE DESIGN. 



The maximum safe tension is taken at one-fourth the ulti- 
mate strength and equals about 16,000 d 2 for cast-steel ropes. 
Three-fourths of this (i2,oood 2 ) may be allowed for bending 
stress and one-fourth (4oood 2 ) for working tension. On this 
basis Tables XXVII-XXIX have been developed. 



Table XXVI. — Values of h for Wire Rope. 

Dry rope on a grooved iron drum 120 

Wet rope on a grooved iron drum 085 

Greasy rope on a grooved iron drum 070 

Dry rope on wood-filled sheaves 235 

Wet rope on wood-filled sheaves 170 

Greasy rope on wood-filled sheaves 140 

Dry rope on rubber and leather filling 495 

Wet rope on rubber and leather filling 400 

Greasy rope on rubber and leather filling 205 



Table XXVII. — Diameters of Minimum Sheaves in Inches. 



Diam. of Rope. 


Steel. 


Iron. 












7-Wire. 


19- Wire. 


7-Wire. 


1 9- Wire. 


1 
4 


24 


14 


48 


28 


ft 


30 


17 


60 


34 


3 

8 


36 


21 


72 


42 


7 
16 


41 


24 - 


83 


48 


1 
2 


47 


28 


94 


56 


ft 


53 


31 


106 


62 


8 


59 


35 


118 


70 


tt 


65 


38 


130 


76 


4 


70 


42 


140 


84 


1 


82 


49 


164 


98 


I 


94 


56 


188 


112 



Table XXVIII. — Deflection of Wire Ropes. 



Def. of still rope at center in feet, 
Def. of driving rope at center in feet, 
Def. of slack rope at center in feet, 
C= span, in feet. 



Steel 
Ao= .000069 C 2 
Ai= .000049 C 2 
A 2 = .000103 C 2 



Iron 
Ao= .000138 C 2 
Ai= .000098 C 2 
A2= .000206 C 3 



BELTS, ROPES, BRAKES, AND CHAINS. 



321 



Table XXIX. — Horse-Powers for a Steel Rope Making a Single Lap on 
Dry, Wood-Filled, Minimum Sheaves. 



Diameter 

of Rope in 

inches. 




Velocity of Rope in 


Ft. per Second. 




10 


20 


30 


40 


50 


60 


1 

4 
5 
16 

3 

2 
9 
T6 
5 
8 
11 
16 
3 
4 
7 

I 


2-5 

3-75 
5-5 
7-5 
9-5 

12 
IS 

18 

21.5 
29 
38 


4-75 
7-5 
10. 5 

14-5 

19 

24 

30 

36 

43 

58 

76 


7 
11 
16 

21-5 

28 
36 
44 
53-5 
63-5 
86.5 
113 


9-5 
14-5 
21 

28.5 

37 

47 

58 

7i 

84 
114 
149 


n-5 

18 

26 

35-5 

46 

58.5 

72 

87-5 
104 
141 
185 


13 -5 

21-5 

30.5 
42 

54-5 
69 

85 
103 
124 
167 
218 



The horse-power that may be transmitted by iron ropes is 
one-half of the above. 

Table XXX (p. 322) gives one manufacturer's standard 
sheave wheel proportions for steel and iron ropes. 

184. Steel Belting.* — The use of flat steel belts, ranging 
from 0.008 to 0.03 inch in thickness, in place of leather belts 
has been a development of recent years. The advantages claimed 
for them are : 

(a) Steel belts need be only from one-half to one-fourth the 
width of leather belts. This means pulleys of narrower face 
and less weight and cost. 

(b) Their own first cost is considerably below that of leather 
or rubber belting. 

(c) They do not stretch or slip after being put on the pulleys 
properly. 

(d) They are not appreciably affected by variations in tem- 
perature or moisture. This makes them very reliable for use 
in damp places. They are especially adapted for use in paint 



* American Machinist, Vol. 37, pp. 852-3. 



322 



MACHINE DESIGN. 




ooooooooooooocoooooonoo >o>fl too t^vo v> 



000\t^OiOl>0 OwO ^-OMOOOi 
•^•00 C5 O\00 ^tN H ro M M M 



O lo W h oo uj r<3 i 



*5K 



tlrOn)roror<)«NHMHHHNNHN 



ifll-cjtlOM N PI HH 



ro ro <N <N in M 



»**. 



HS 



■*-S|2-N-|2HS- 1 --|5HSH-asS;**^!-i-t5-i-H-35 



5^t't'^-'^-'^-f0r0(NCMMMMMP)MMI 



lOO m lO Ifl io xf ■* ro c: M M (M M fON <n ro<N 



lO^t'^fOrO'^-rOrON <N h h i 



t-O O io lo >o io "^- ro fO ' 



O00 O\00 00 00 00 O loiorooo CN 00 IOCS O -<tO 



rf <N M O <^00 t^vO lO'tfON PI H flUJN ft N 



OhOhmmmOOO OvO OO w 



rf w w O 0\ f» l>0 iO"*fON<NHr<5<NMM 



BELTS, ROPES, BRAKES, AND CHAINS. 



323 



or varnish works, etc., as they can be washed off readily with 
gasoline. 

On the other hand, they are more sensitive than other belts, 
and shafts and pulleys must be in line and level or the belt will 
run to the low side of the pulley and may run off. Crown pulleys 
cannot be used for them in any case. They will run on flat- 
faced, uncovered, iron or steel pulleys, as well as on wood pulleys, 
but the use of canvas or rubber pulley covering is so beneficial 
that it seems almost necessary to good service. 

185. Block and Band Brakes. — For the sake of unity of 
treatment block brakes will be considered at this point. The 
formula? are all developed by equating to zero the sum of the 
moments about the axis of the brake lever, considering the latter 
as a free body. 

F = force in pounds at end of brake lever; 

P = tangential force in pounds at rim of brake wheel, called 
braking force ; 

/i = coefficient of friction between brake block and brake wheel. 

1. Block brake, Fig. 186-B. For rotation in 



either direction: 



F = 



Pb 

a + b 




2. Block brake, Fig. 186C. Upper sign 
for clockwise, lower sign for counter-clockwise 
rotation : 

a + b\n b. 

3. Block brake, Fig. 1S6D. Upper sign 
for clockwise, lower sign for counter-clockwise 
rotation : 

Pb 
a+b\fi 



Fig. iS6B- 



F = 



;(H)- 




3 2 4 



MACHINE DESIGN. 



The brake wheel and friction-block may be grooved as shown 
in Fig. 186E. In this case substitute for p. in the foregoing 




equations the value 



, where a is 



Fig. 186E. 



sina+/i cos a 

one -half the angle included by the faces of the 

grooves. 

The formulae for band brakes, simple and 

differential, are developed from the belt for- 

T 
mula, log e — r =fi.0j which may be written — = e fid ; e being the base 
J- 2 J- 2 

of natural logarithms = 2.71828. 6 is the angle of contact of 
brake and drum in radians. 

/j. is the coefficient of friction. 

P = tangential force in pounds at rim of brake drum; 

F = force in pounds at end of brake lever; 

T\ = tension in tight side of band, pounds; 

T2 = tension in slack side of band, pounds. 

T X -T^P, 



Ti = P- 



e^ 



To = P 



1. Simple band brake, Fig. 1S6F, 

Pbl e^ 
For clockwise rotation: F = — ( — 2 — 

a V 119 - 



For counter-clockwise rotation: F 



_Pb/ 



2. Simple band brake, Fig. 186G. 
For clockwise rotation: F = — — a ). 



For counter-clockwise rotation : F = — 

a 




a V* 



Fig. 186G. 



BELTS, ROPES, BRAKES, AND CHAINS. 325 

3. Differential band brake, Fig. 186H. 

P/b 2 e^-h 

For clockwise rotation : F = - 



a \ er -1 
P 



For counter-clockwise rotation: F = 



P( b 2 -he» e \ 
a\ e» d -i )' 



In this case if b 2 is equal to or less than bid*, F will be zero 
or negative and the brake sets automatically. 





*** 



Fig. 1S6H. Fig. 186/. 

4. Differential band brake, Fig. 186/. 

For clockwise rotation: i* = — I — 5 

.11- ♦ ♦■ 77 P(b^ e + b 2 
For counter-clockwise rotation : b = 



If bi = b 2 = b, for both cases: i 7 



Pfr/ ^ + A 
a\e"-i)' 



186. Chains. — Various types of chains are used for power 
transmission purposes. 

(a) Tests made at the University of Illinois * show that 
the ordinary method of computing round-rod chain strength, 
both open and stud link, is incorrect. They show that a load 
P on the link, while it produces an average intensity of stress 

P 

in the cross-section containing the minor axis = — , a being the 

* Bulletin No. 18. 



326 



MACHINE DESIGN. 



Ttd? 
area of the rod = — , produces a much greater maximum fiber 
4 

stress than this. With an open link of usual proportions the 

maximum tensile stress is approximately four times this value, 

2 p 

The introduction of a stud in the link equalizes the 



or 



stresses somewhat throughout the link and reduces the maximum 
tensile stress about 20 per cent. The following formulae are 
applicable to chains of the usual form: 

P = o.4d 2 f t , for open links; 
P = o.$d 2 f t , for stud links; 

f t = allowable unit tensile stress = 15,000 to 20,000 (maxi- 
mum) pounds per square inch. 





Fig. 186/. 



Fig. i862L 



(b) Flat link chains of either the block or roller type, Figs. 
186/ and K, are used for power transmission purposes. They 
are not suitable for high speeds. Not only do they tend to wear 
and stretch, which throws them out of equality of pitch with their 
sprockets, but from the nature of their construction it is inev- 
itable that they transmit motion at a continuously varying velocity 
ratio. They have a wide field of usefulness, however, for low- 
speed transmission where an absolutely uniform velocity ratio 
is not required nor noise objectionable. Their proportioning 
involves two chief points, the shearing strength of the pins and 
the tensile strength of the side plates at their minimum section. 
If kept free from dust, well-lubricated, not overloaded and not 
run at too high a speed, they show efficiencies ranging as high as 
94 per cent. 



BELTS, ROPES, BRAKES, AND CHAINS. 



327 



(c) Some of the objections to flat link chains, particularly 
those connected with change of pitch of chain and sprocket 
with stretch and wear, have been met and conquered in so-called 
silent, high-speed chains, such as the Renold and Morse. These, 
because they continue to lit their sprockets, run with much less 




Fig. 186L. 



noise than block or roller chains, may be run at considerably 
higher speeds, will transmit much greater powers, and when 
properly installed and cared for show extremely high efficiencies. 

Fig. 186L, from the Morse Chain Company's catalog, shows 
the action of this chain and clearly illustrates their employment 
of the extremely efficient knife-edge bearing at the joints. 



CHAPTER XVI. 

FLY-WHEELS AND PULLEYS. 

187. Theory of Fly-wheel. — Often in machines there is 
capacity for uniform effort, but the resistance fluctuates. In 
other cases a fluctuating effort is applied to overcome a uniform 
resistance, and yet in both cases a more or less uniform rate 
of motion must be maintained. When this occurs, as has been 
explained,* a moving body of considerable weight is interposed 
between effort and resistance, which, because of its weight, 
absorbs and stores up energy with increase of velocity when 
the effort is in excess, and gives it out with decrease of velocity 
when the resistance is in excess. This moving body is usually 
a rotating body called a fly-wheel. 

To fulfill its office a fly-wheel must have a variation of 
velocity, because it is by reason of this variation that it is able 
to store and give out energy. The kinetic energy, E, of a 
body whose weight is W lbs., moving with a velocity v feet per 
second, is expressed by the equation 

-_ Wv 2 
hi = . 

To change E, with W constant, v must vary. The allowable 
variation of velocity depends upon the work to be accomplished. 
Thus the variation in an engine running electric lights or spin- 
ning-machinery should be very small, probably not greater 

* See § 43. 

328 



FLY- W HEELS AND PULLEYS. 329 

than a half of one per cent, while a pump or a punching- 
machine may have a much greater variation without interfering 
with the desired result. If the maximum velocity, v\, of the 
fly-wheel rim and the allowable variation are known, the mini- 
mum velocity, v 2 , becomes known; and the energy that can be 
stored and given out with the allowable change of velocity is 
equal to the difference of kinetic energy at the two velocities. 

Wv x 2 Wv 2 2 W 
JE = = — (vi 2 -v 2 2 ). 

2g 2g 2g 

188. The general method for fly-wheel design is as follows: 
Find the maximum energy due to excess or deficiency of effort 
during a cycle of action, = J£. Use the foot-pound-second 
system of units. Assume a convenient mean diameter of fly- 
wheel rim. From this and the given maximum rotative speed 
of the fly-wheel shaft find V\. Solve the above equation for 
W thus: 

2gAE 



W 



Vi 2 -v 2 2 



Substitute the values of JE, v x , v 2 , and # = 32.2 feet per second 2 , 
whence W becomes known, = weight of fly-wheel rim. The 
weight of rim only will be considered; the other parts of the 
wheel, being nearer the axis, have less velocity and less capacity 
per pound for storing energy. Their effect is to reduce slightly 
the allowable variation of velocity.* 

189. Problem. — In a punching-machine the belt is capable 
of applying a uniform torsional effort to the shaft; but most 



* Numerical examples taken from ordinary medium-sized steam-engine fly- 
wheels show that while the combined weight of arms and hub equals about one 
third of the total weight of the wheel, the energy stored in them for a given varia- 
tion of velocity is only about 10 per cent of that stored in the rim for the same 
variation. 



33° MACHINE DESIGN. 

of the time it is only required to drive the moving parts of the 
machine against frictional resistance. At intervals, however, 
the punch must be forced through metal which offers shearing 
resistance to its action. Either the belt or fly-wheel, or the two 
combined, must be capable of overcoming this resistance. A 
punch makes 30 strokes per minute, and enters the die J inch. It 
is required to punch f-inch holes in steel plates \ inch thick. 
The shearing strength of the steel is about 50,000 lbs. per square 
inch. When the punch just touches the plate the surface which 
offers shearing resistance to its action equals the surface of the 
hole which results from the punching, = ndt, in which d = diam- 
eter of hole or punch, / = thickness of plate. The maximum 
shearing resistance, therefore, equals TrfXjX 50,000 = 58,900 lbs. 
As the punch advances through the plate the resistance decreases 
because the surface in shear decreases, and when the punch 
just passes through the plate the resistance becomes zero. If 
the change of resistance be assumed uniform (which would 
probably be approximately true) the mean resistance to punching 
would equal the maximum resistance + minimum resistance, ^2, 

= - =29,450. The radius of the crank which actuates 

the punch = 2 in. In Fig. 187 the circle represents the path of 
the crank-pin center. Its vertical diameter then represents the 
travel of the punch. If the actuating mechanism be a slotted 
cross-head, as is usual, it is a case of harmonic motion, and it may 
be assumed that while the punch travels vertically from A to B, 
the crank-pin center travels in the semicircle ACB. Let BD 
and DE each = \ inch. Then when the punch reaches E it 
just touches the plate to be punched, which is \ inch thick, and 
when it reaches D it has just passed through the plate. Draw the 
horizontal lines EF and DG and the radial lines OG and OF. 
Then, while the punch passes through the plate, the crank-pin 
center moves from F to G, or through an angle (in this case) 



FLY- WHEELS AND PULLEYS. 



331 



of 1 9 . Therefore the crank-shaft A, Fig. 188, and attached 

gear rotate through 19 during the action of the punch. The 

ratio of angular velocity of the pinion and the gear = the inverse 

60 
ratio of pitch diameters = — =5. 



Hence the shaft B rotates 



12 



through an angle =i9°X5 =95° during the action of the punch. 
If there were no fly-wheel the belt would need to be designed 
to overcome the maximum resistance; i.e., the resistance at the 
instant when the punch is just beginning to act. This would 




Fig. 187. 



Tt 



Crank 



W 



! 

SO 

CO 



Fig. ii 



give for this case a double belt about 20 inches wide. The need 
for a fly-wheel is therefore apparent. Assume that the fly- 
wheel may be conveniently 36 inches mean diameter, and that 
a single belt 5 inches wide is to be used. The allowable maxi- 
mum tension is then =5Xallowable tension per inch of width 
of single belting = 5 X70 = 350 lbs. = T X . 

Since the pulley-shaft makes 150 revolutions per minute and 
the diameter of the pulley is 2 feet, the velocity of the belt = 
150X2X71 = 942 feet per minute. At this slow speed the simple 

T 
form of the belt formula may be used, i.e., log tjt= 0.4343 fid. 

-L 2 

Assume an angle of contact of 180 . Then 



33 2 MACHINE DESIGN. 

0=3.1416, 

T 
log ^-=0.4081; 
-i 2 

and .-. — 1 = 2.56. 

i 2 

r 2 =g-= 13 6. 7 ibs. 

Ti — T 2 = 213.3 lbs. = the driving force at the surface of the pulley. 
Assume that the frictional resistance of the machine is 
equivalent to 25 lbs. applied at the pulley-rim. Then the belt 
can exert 213.3 — 25=188.3 lbs. =P, to accelerate the fly-wheel 
or to do the work of punching. Assume variation of velocity =10 
per cent. The work of punching = the mean resistance offered 
to the punch multiplied by the space through which the punch 

^8000 
acts, = ;d - z — xo-5" = 14725 in. -lbs. =1225 ft. -lbs. The pulley- 
shaft moves during the punching through 95 , and the driv- 
ing tension of the belt, =P = 188.3 ^s., does work=PXspace 

05° 
moved through during the punching = 188.3 lbs. X 7^-7- = 188.3 

300 

lbs.X7iX2 ft. X-^r- =312 ft.-lbs. The work left for the fly- 
wheel to give out with a reduction of velocity of 10 per cent 
= 1225—312=913 ft.-lbs. Let v\ = maximum velocity of fly- 
wheel rim; v 2 = minimum velocity of fly-wheel rim; W = weight 
of the fly-wheel rim. The energy it is capable of giving out, 

W(v 1 2 -v 2 2 ) 
while its velocity is reduced from Vi to v 2 , = , and 

the value of W must be such that this energy given out shall 
equal 913 ft.-lbs. Hence the following equation may be written: 

W(v 1 2 ~v 2 2 ) 
=013. 



FLY-WHEELS AND PULLEYS. 333 

Therefore w=^^. 

v 1 2 —v 2 2 

The punch-shaft makes 30 revolutions per minute and the pulley- 
shaft 30X5 = 150 =iV revolutions per minute. Hence v x in 

NDx 

feet per second= —7 — , D being fly-wheel diameter in feet =3 feet. 

150X371 
*i=—^— =23.56; 

V2 =0.90^1 =21.2; 
^i 2 =5555 ^2 2 =449; ^i 2 -^2 2 = io6. 

Hence FF= 9 -^f^ =555 «*• 

To proportion the rim: A cubic inch of cast iron weighs 

0.26 lb.; hence there must be 7 = 2135 cu. ins. The cubic 

' 0.26 0D 

contents of the rim=mean diameter XttX its cross-sectional area 

A =2135 cu. ins.; hence 

„ 2I 35 



36" Xt: 



10.45 so i- ms - 



If the cross-section were made square its side would =v 18.45 

= 4-3- 

190. Pump Fly-wheel. — The belt for the pump, p. 304, is 
designed for the average work. A fly-wheel is necessary to 
adapt the varying resistance to the capacity of the belt. The 
rate of doing work on the return stroke (supposing no resistance 
due to suction) is only equal to the frictional resistance of the 
machine. During the working stroke the rate of doing work 
varies because the velocity of the plunger varies, although the 
pressure is constant. The rate of doing work is a maximum 
when the velocity of the plunger is greatest. In Fig. 189, A 
is the velocity diagram, B is the force diagram, C is the 



334 



MACHINE DESIGN. 



tangential diagram drawn as indicated on pp. 78-80. The 
belt, 5 inches wide, is capable of applying a tangential force 
of 148 lbs. to the 18-inch pulley- rim. The velocity of the pulley- 
rim = 7ri. 5 X 300 = 1414/ . The velocity of the crank-pin axis 
= 71X0.833X50 = 130.9/. Therefore the force of 148 lbs. at 



1414 



= 1599 lbs - 



the pullev-rim corresponds to a force =i48X 

' 130-9 

applied tangentially at the crank-pin axis. This may be plotted 
as an ordinate upon the tangential diagram C, from the base 





Fig. iJ 



line XX 1, using the same force scale. Through the upper 
extremity of this ordinate draw the horizontal line DE. The 
area between DE and XX X represents the work the belt is 
capable of doing during the working stroke. During the return 
stroke it is capable of doing the same amount of work. But 
this work must now be absorbed in accelerating the fly-wheel. 
Suppose the plunger to be moving in the direction shown by 
the arrow. From E to F the effort is in excess and the fly-wheel 
is storing energy. From F to G the resistance is in excess and 
the fly-wheel is giving out energy. The work the fly-wheel must 
be capable of giving out with the allowable reduction of velocity 
is that represented by the area under the curve above the line 
FG. From G to D, and during the entire return stroke, the 
belt is doing work to accelerate the fly-wheel. This work 



FLY- W 'HEELS AND PULLEYS. 335 

becomes stored kinetic energy in the fly-wheel. Obviously the 
following equation of areas may be written : 

XiEF+XGD + XHKXi =GMF. 

The left-hand member of this equation represents the work 

done by the belt in accelerating the fly-wheel; the right-hand 

member represents the work given out by the fly-wheel to help 

the belt. 

The work in foot-pounds represented by the area GMF 

may be equated with the difference of kinetic energy of the 

fly-wheel at maximum and minimum velocities. To find the 

value of this work: One inch of ordinate on the force diagram 

represents 8520 lbs.; one inch of abscissa represents 0.449 foot. 

Therefore one square inch of area represents 8520 lbs. X 0.449' 

= 3825.48 ft. -lbs. The area GMF =0.4 sq. ins. Therefore 

the work = 3825.48X0.4 = 1530 ft. -lbs. = JE. The difference of 

W 
kinetic energy =— (vi 2 — v 2 2 ) = 1530; W equals the weight of 

the fly-wheel rim. Hence 

1530X32.2X2 



W 



Vi 2 —V 2 2 



Assume the mean fly-wheel diameter = 2.5 feet. It will be 
keyed to the pulley-shaft, and will run 300 revolutions per 
minute, = 5 revolutions per second. The maximum velocity 
of fly-wheel rim =71X2.5 X 5 =39.27 =^1. Assume an allowable 
variation of velocity, =5 per cent. Then i>2 = 37-2 7 X 0.95 = 
37.3; i/i 2 = 1542.3; V2-1391.3; v 1 2 -v 2 2 =i$i. Hence 

iSyX32.2X 8 
I 5 I 

There must be 651 -^0.26 cu. in. in the rim, =2504. The mean 
circumference =30" X7r = 94.2". Hence the cross-sectional area 



336 MACHINE DESIGN. 

of rim =2504-^94.2 =26.6 sq. ins. The rim may be made 
4.5" X 6". 

The frictional resistance of the machine is neglected. It 
might have been estimated and introduced into the problem as a 
constant resistance. 

191. Steam-engine Fly-wheel. — From given data draw the 
indicator-card as modified by the acceleration of reciprocating 
parts. See page 77 and Fig. 46. From this and the velocity 
diagram construct the diagram of tangential driving force, 
Fig. 47. Measure the area of this diagram and draw the equiva- 
lent rectangle on the same base. This rectangle represents the 
energy of the uniform resistance during one stroke; while the 
tangential diagram represents the work done by the steam upon 
the crank-pin. The area of the tangential diagram which 
extends above the rectangle represents the work to be absorbed 
by the fly-wheel with the allowable variation of velocity.* Find 
the value of this in foot-pounds, and equate it to the expression 
for difference of kinetic energy at maximum and minimum 
velocity. Solve for W, the weight of fly-wheel. 

192. Stresses in Fly-wheel Rims. — Mathematical analyses 
of the stresses in fly-wheel rims are unsatisfactory. In the first 
place, in order to get solutions of reasonable simplicity it is 
customary to make assumptions which are contrary to the 
actual conditions; and in the second place, no satisfactory 
data exist concerning the strength of cast iron in such heavy 
sections as are used in large engine fly-wheels. An examination 
of the nature of the stresses, however, will indicate the points 
to be looked out for in design. 

Considering a ring of hollow cylindrical form, comparatively 

* For compound engines and for varying resistances the dia grams should be 
constructed for the complete cycle. For full treatment of the problem of fly 
wheels for engines driving alternators the reader is referred to the Trans. A. S. 
M. E., Vol. XXII, p. 955, and Vol. XXIV, p. 98. 




FLY- WHEELS AND PULLEYS. 337 

thin radially, it can be shown that, when it is rotated about its 
axis, tension is set up in the ring proportional to the weight of 
the material used and the square of the linear velocity. This 
tension is due solely to the action of "centrifugal force" and is 
termed "centrifugal tension." 

Consider the half- ring shown in Fig. 190: 
i/=the velocity of the rim in feet per second. 
c = " centrifugal force" per foot of rim; 
R= radius in feet; 
A =area of rim in square inches; 
P=total tension in rim in pounds; 
ft =unit tensile stress in rim in pounds; 
w= weight of material as represented by a piece 1 inch 

square and 1 foot long; 
g =32.2 feet per second per second; 
2P=sum of horizontal components of all the small centri- 
fugal forces cds; 
Each horizontal component =cds cos d, which may be written 
cR cos dd, because ds =Rdd. 

IT 

2P= f 2 cR cos ddd = 2cR; 



-x: 



But. c = 



/. P = cR. 

Mv 2 W v 2 



R g R 1 
Wv 2 

■'• T' 

W being the weight of one linear foot of rim = wA 

Also, P=M; 

r A W „ wAv 2 



g g 






338 MACHINE DESIGN. 

For cast iron, putting f t = 20,000 lbs., the ultimate strength, 
and ^ = 0.26X12 lbs., it follows that 1^ = 454 feet per second. 
In other words a cast-iron ring will burst at a speed of 454 feet 
per second. Furthermore, an examination of the formula 
shows that for a ring this bursting velocity depends not at all 
on the size or shape of the cross-section, but only on the material 
used as represented by f t and w. This is not entirely true. In 
a cylindrical disk, without any hole, the maximum centrifugal 

0.41WV 2 
tension is at the center, where f t = . For a cylindrical 

o 

0.&2WV 2 

disk with incipient hole, f t =— . For the derivation of these 

o 

formula?, see Ewing's Strength of Materials; also Stodola's Steam 
Turbines* 

This centrifugal tension causes a corresponding elongation of 
the material and therefore an increase in the radius of the ring. A 
free, thin ring of whatever cross-section can and does take the new 
radius and the tension on all sections- =/« pounds per square inch. 

With the introduction of rigidly fastened arms a number of 
new and vital elements enter into the problem. An arm of the 
same original length as the original radius of the rim when 
rotated about an axis perpendicular to its inner end will also 
suffer an elongation due to centrifugal action. The amount of this 
radial elongation will vary with the form of the arm, but in no 
practical case will it amount to as much as one third of the 
radial increase of the ring rotating at the same speed. 

To accommodate this difference the arm, if rigidly fastened 
to hub and rim, will be extended lengthwise by the rim and the 
rim will be drawn in, out of its regular circular form, by the 
arm. The relation between the amount the arm is drawn out 
and the amount the rim is drawn in is governed by the propor- 
tions of these parts. 

* See also, Moss in Trans. A. S. M. E., Vol. XXXIV. 



FLY-IVHEELS AND PULLEYS. 339 

The result is that the rim tends to bow out between the arms 
and really become akin to a uniformly loaded continuous beam 
with the dangerous sections midway between the arms and at 
the points of junction of arms and rim. The fallacy of applying 
the ring theory solely to the fly-wheel rim becomes evident at 
once. In a free ring the form of cross-section is immaterial, as 
the section is subjected only to tension. In the rim with arms 
the form of cross-section becomes a vital point, as the rim is 
subjected to flexure as well as tension, and the strength of a 
member to resist flexure depends directly upon the modulus 
of the section. 

In addition to the foregoing stresses, which are induced 
under all conditions, even under the extreme supposition that 
the wheel is rotating at a perfectly uniform rate, there are others 
when the rim is considered as performing its functions — i.e., 
in a balance-wheel, absorbing or giving out energy by changes 
of velocity and, in a band-wheel, transmitting the power. 

This may be seen by reference to Fig. 191. A shows the 
relation between rim, arm, and hub when the wheel is at rest 
or rotating uniformly and not transmitting 
any power. B shows the relation when 
work is being done. The arm becomes an 
encastre beam and corresponding stresses 
are induced in it. Furthermore, the bend- 
ing of the arm tends to shorten it radially, thus drawing in the 
outer end, which increases the flexure in the rim. In addition 
to the foregoing there are stresses in the rim due to the weight 
of the wheel, shrinkage, etc., which cannot be eliminated. 

193. Stresses in Arms of Pulleys or Fly-wheels. — The arms 
are principally stressed by the bending moment due to variations 
of velocity of the wheel or to the power transmitted. 

Let M t = the greatest turning moment transmitted in inch- 
pounds; 




34° MACHINE DESIGN. 

n = number of arms; 

ft = safe unit stress in outer fiber of arm in pounds per 
square inch; 

— = modulus of section of arm, dimensions in inches. 

Then 

M t = nf t — may be written and solved for—. 
c c 

Having determined upon the form of cross-section the dimensions 

can be determined from this value of—. 

c 

If M t is unknown the arms can be made as strong as the shaft 

by equating the twisting strength of the shaft to the bending 

strength of the arms, thus : 

7zr 3 / 

f 8 = allowable shearing stress in outer fiber of shaft, pounds per 

square inch; 
r = radius of shaft in inches; 

#n, ft, and — as before. 
C 
Consider junction of arm and hub next. (See 
Fig. 102.) 
Fig. 192. ° y 

The tendency for the arm to fail through 

flexure on the section A-A may be equated to the tendency 
for the bolts 2 and 3 to shear off, using 1 as a pivot. 

Let A = combined shearing areas of 2 and 3, square inches; 
f s = allowable shearing stress of 2 and 3 in pounds per 

square inch; 
/ = distance between centers 1 and 2, and 1 and 3, in 
inches; 



Then 



FLY-IVHEELS AND PULLEYS. 341 

jt = allowable stress in outer fiber of arm in pounds per 
square inch; 

— = modulus of arm section, dimensions in inches. 
c 



AUi = U-> 



which can be solved for A, the desired area. 

If the arm is bolted to the rim a similar method may be 
employed to make the bolts as strong as the arm. 

194. Construction of Fly-wheels. — Since weight is so great 
a factor in fly-wheels it has been common practice to make 
them of that material which combines greatest weight with 
least cost, namely, cast iron. That this is not always safe 
practice has been conclusively demonstrated by many serious, 
accidents. 

Up to 10 feet in diameter the wheels are generally cast 
in a single piece. Occasionally the hub is divided to relieve 
the stresses due to cooling. In such cases, supposing the wheel 
to have six arms, the hub is made in three sections, each having 
a pair of arms running to the rim. Since the sections are inde- 
pendent, any pair of arms can adjust itself to the conditions ci 
shrinkage without subjecting the other arms to indeterminate 
stresses. The hub sections are separated from each other by 
a space of half an inch or less and this is filled with lead 01 
babbitt metal. Then shrink-rings or bolts are used to hold 
the sections together. Sometimes the hub is only split into two 
parts. 

For reasons connected chiefly with transportation, wheels 
from 10 to 15 feet in diameter are cast in two halves which 
are afterwards joined together by flanges and bolts at the rim, 
and shrink-rings or bolts at the hub. 



342 MACHINE DESIGN. 

In still larger and heavier wheels the hub is generally made 
entirely separate from the arms. The rim is made in as many 
segments as there are arms. Sometimes the arm is cast with 
the segment and sometimes the arms and segments are cast 
separately. The hub is commonly made in the form of a pair 
of disks having a space between them to receive the arms which 
are fastened to them by means of accurately fitted through bolts. 

Unless the wheel is to be a forced fit on its shaft it is best 
to have three equally spaced key ways, so that it may be kept 
accurately centered with the shaft. 

In these large wheels the joints of the segments of the rim 
are usually midway between the arms and steel straps or links 
C^C-^Tl r~^~S=^r^/ sucn as are shown in Fig. 193 are 
. U — jf ->•/ U-^^-J heated and dropped into recesses pre- 
FlG * I93 ' viously fitted to receive them. As they 

cool, their contraction draws the joint together. They should 
not, however, be subjected to a very great initial tension of this 
sort. The form shown at A is most commonly used. The links 
are made of high-grade steel and their area is such that their ten- 
sile strength equals that of the reduced section of the rim. The 
areas subjected to shear and compression must also have this 
strength. 

Taking the nature of the stresses into consideration it is clear 
that the rim should always be as deep radially as possible to 
resist the flexure action, also that the arms should be near together. 
Many arms are much better than a few and a disk or web is still 
better.* 

The strongest wheel having arms will be one whose rim is 
cast in a single piece, while the arms and hubs are cast as a second 
piece. On the inside of the rim there are lugs between which 

* Disk wheels have the further advantage of offering less resistance to the 
air. This may be a considerable item. See Cassier's Mag., Vol. 23, pp. 577 and 
761. 




FLY-IVHEELS AND PULLEYS 343 

the ends of the arms fit so that there is a space of about one 
fourth inch all around. (See Fig. 194.) This space may be 
filled with oakum well driven in. It is clear that the rim in this 
case acts as a free ring and is subjected solely to centrifugal 
tension.* 

Joints in the rim must always be a source of weakness whether 
located at the end of the arms or midway between arms. 

Tension 
Rod 

Fig. 194. Fig. 195. 

If a flanged joint midway between the arms is used, such as 
is shown in Fig. 195, which is common practice for split band- 
wheels of medium size, the flanges should be deep radially and 
well braced by ribs. The bolts should be as close to the rim 
as possible, and a tension rod should carry the extra stress (due 
to the weight of the heavy joint and its velocity) to the hub. 
Experiments made by Prof. C. H. Benjamin f show that the use 
of such tie-rods increases the strength of the wheel 100 per cent 
over that of a similar wheel without tie-rods. He also found 
that jointed rims are only one fourth as strong as solid rims. 

Probably as strong a form of cast-iron built-up wheel for 
heavy duty as any yet designed is one described by Mr. John 
Fritz,| having a hollow rim and many arms. 

But, at the best, cast iron is an uncertain material to use 
for such tensile and flexure stresses as are induced in a heavy- 
duty fly-wheel, and it is wiser to make such wheels of structural 
steel. A built-up wheel having a disk or web of steel plates and 
a rim of the same material, all joints being carefully " broken " 
and strongly riveted, is so much better than any built-up cast- 

* See Trans. A. S. M. E., Vol. XX, p. 944, and Vol. XXI, p. 322. 
t Ibid., Vols. XX and XXIII. 
X Ibid., Vol. XXI. 



344 MACHINE DESIGN. 

iron wheel that the latter are passing out of use. The steel 
wheels can have at least twice the rim velocity of the cast wheels 
with greater safety and may therefore be much lighter for the 
same duty. Their lesser weight makes less pressure on the 
bearings and consequently less friction loss.* 

Plate II shows forms of rim joints for split rim flywheels 
and pulleys which are probably as strong as any that can be 
devised. They are taken from the American Machinist, Vol. 30. 
It is a mistake, however, to believe that any of these joints will 
will give as strong a wheel as one having a solid rim. 

Pulley rims, according to Carl Barth, should have a mini- 
mum width, F, equal to 1-^- times the belt width + 3%- inch. Their 
thickness is a question of sound castings and avoidance of shrink- 
age stresses rather than of strength which can be computed. 
The minimum finished thickness at the edge of pulleys 6 or 8 
inches in diameter may be as little as f inch, rising to f inch 
for 72 inches diameter. The radial height of crown may be 

— . The diameter of hub may be made twice the bore. Six 

arms of elliptical cross-section are most frequently employed, 
the minor axis being about one-half the major. To get the width 
of the arm at the hub, the circumference of the latter is divided 
into as many equal parts as there are arms. The arms are 
tapered in width, about \ inch per foot, but may have uniform 
thickness. Generous fillets should be employed where they 
join each other and the hub, as well as where they join the rim. 
Transition from thick to thin sections must be made as gradual 
as possible. For extra width of face two parallel sets of arms 
may be used. 



* For drawings and descriptions of wheels made of forged materials the reader 
is referred to Vol. XVII, Trans. A. S. M. E., and Power, April 1894, Nov. 1895, 
Jan. 1896, and Nov. 1897. Also, Jones' Machine Design. 



Plate II. 




HAIGHT'S JOINT FOR HEAVY RIM. 



FLANGED JOINT OVER_ARM 
SEGMENTAL RIM. 



i<§)^(6) O 




DOUBLE ARM JOINT FOR WIDE RIM. 





\o u o u o 



DOUBLE ARM JOINT FOR SHEAVE WHEEL. 



345 



CHAPTER XVII. 

TOOTHED WHEELS OR GEARS. 

195. Fundamental Theory of Gear Transmission. — When 
toothed wheels are used to communicate motion, the motion 
elements are the tooth surfaces. The contact of these surfaces 
with each other is line contact. Such pairs of motion elements 
are called higher pairs, to distinguish them from lower pairs, 
which are in contact throughout their entire surface. Fig. 196 
shows the simplest toothed-wheel mechanism. There are three 
links, a, b, and c, and therefore three centres, ab, be, and ac. These 
centres must, as heretofore explained, lie in the same straight 
line, ac and ab are the centers of the turning pairs connecting 
c and b to a. It is required to locate be on the line of centers. 

When the gear c is caused to rotate uniformly with a certain 

angular velocity, i.e., at the rate of m revolutions per minute, 

it is required to cause the gear b to rotate uniformly at a rate 

of n revolutions per minute. The angular velocity ratio is there- 

tn 
fore constant and = — . The centre be is a point on the line of 

centers which has the same linear velocity whether it is con- 
sidered as a point in b or c. The linear velocity of this point be 
in b = 2~Rin; and the linear velocity of the same point in c=27zR 2 m ; 
in which Ri = radius of be in b, and i? 2 =radius of be in c. But 
this linear velocity must be the same in both cases, and hence the 
above expressions may be equated thus: 

2izR\n = 2nR2tn, 

347 



3*8 
whence 



MACHINE DESIGN. 



*1 

R2 



Hence be is located by dividing the line of centers into parts which 

are to each other inversely as the angular velocities of the gears. 

Thus, let ab and ac, Fig. 197, be the centers of a pair of gears 

m 
whose angular velocity ratio =— . Draw the line of centers; 

n 7 

divide it into m+n equal parts; m of these from ab toward the 
right, or n from ac toward the left, will locate be. Draw circles 
through be, with ab and ac as centers. These circles are the 
centrodes of be and are called pitch circles. It has been already 
explained that any motion may be reproduced by rolling the 
centrodes of that motion upon each other without slipping. 




Fig. 196. 



Fig. 197. 



Therefore the motion of gears is the same as that which would 
result from the rolling together of the pitch circles (or cylinders) 
without slipping. In fact, these pitch cylinders themselves 
might be, and sometimes are, used for transmitting motion of 
rotation. Slipping, however, is apt to occur, and hence these 
"friction-gears" cannot be used if no variation from the given 
velocity ratio is allowable. Hence teeth are formed on the 
wheels which engage with each other, to prevent slipping. 

196. Definitions. — If the pitch circle be divided into as many 
equal parts as there are teeth in the gear, the arc included between 



TOOTHED WHEELS OR GEARS. 349 

two of these divisions is the circular pitch * of the gear. Circular 
pitch may also be defined as 'the distance on the pitch circle 
occupied by a tooth and a space; or, otherwise, it is the distance 
on the pitch circle from any point of a tooth to the corresponding 
point in the next tooth. A fractional tooth is impossible, and 
therefore the circular pitch must be such a value that the pitch 
circumference is divisible by it. Let P = circular pitch in inches; 
let D= pitch diameter in inches; iV = number ' of teeth; then 

NP=t:D\ N = ~; D=~ — ; P=^ry. From these relations 

r 71 iv 

any one of the three values, P, D, and N, may be found if the 

other two are given. 

Diametral pitch is the number of teeth per inch of pitch 

N 
diameter. Thus if ^= diametral pitch, ^=77- Multiplying the 

n nD , N , „ tzD N 

two expressions, P = ~j^r and p =— , together gives Pp =~fj . — =7r. 

Or, the product of diametral and circular pitch =n. Circular 
pitch is usually used for large cast gears, and for mortise-gears 
(gears with wooden teeth inserted). Diametral pitch is usually 
used for small cut gears. 

In Fig. 198, b, c, and k are pitch points of the teeth; the arc 
bk is the circular pitch) ab is the face of the tooth; bm is the flank 
of the tooth ; the whole curve abm is the profile of 
the tooth; AD is the total depth of the tooth; AC 
is the working depth; AB is the addendum; a 
circle through A is the addendum circle. Clear- 
ance is the excess of total depth over working FlG * I98 ' 
depth, = CD. Backlash is the width of space on the pitch lire 
minus the width of the tooth on the same line. In cast gears 
whose tooth surfaces are not "tooled," backlash needs to be 




* Sometimes called circumjerential pitch. 



350 MACHINE DESIGN. 

allowed, because of unavoidable imperfections in the surfaces. 
In cut gears, however, it may be reduced almost to zero, and the 
tooth and space, measured on the pitch circle, may be considered 
equal. 

197. Conditions Governing Forms of Teeth. — Teeth of almost 
any form may be used, and the average velocity will be right. 
But if the forms are not correct there will be continual variations 
of velocity ratio between a minimum and maximum value. These 
variations are in many cases unallowable, and in all cases unde- 
sirable. It is necessary therefore to study tooth outlines which 
shall serve for the transmission of a constant velocity ratio. 

The centro of relative motion of the two gears must remain 
in a constant position in order that the velocity ratio shall be 
constant. The essential condition for constant velocity ratio is, 
therefore, that the position of the centro of relative- motion of the 
gears shall remain unchanged. If A and B, Fig. 199, are tooth 
surfaces in contact at a, their only possible relative motion, if 
they remain in contact, is slipping motion along the tangent CD. 
The centro of this motion must be in EF, a normal to the tooth 
surfaces at the point of contact. If these be supposed to be 
teeth of a pair of gears, b and c, whose required velocity ratio 
is known, and whose centro, be, is therefore located, then in 
order that the motion communicated from one gear to the other 
through the point of contact, a, shall be the required motion, it 
is necessary that the centro of the relative motion of the teeth 
shall coincide with be. 

198. Illustration. — In Fig. 200, let ac and ab be centers of 
rotation of bodies b and c, and the required velocity ratio is such 
that the centro of b and c falls at be. Contact between b and c 
is at p. The only possible relative motion if these surfaces re- 
main in contact is slipping along CD', hence the centro of this 
motion must be on EF, the normal to the tooth surfaces at the 
point of contact. But it must also be on the same straight line 



TOOTHED WHEELS OR GEARS. 



35t 



with ac and ab; hence it is at be, and the motion transmitted for 
the instant, at the point p, is the required motion, because its 
centro is at be. But the curves touching at p might be of such 
form that their common normal at p would intersect the line of 
centers at some other point, as K, which would then become the 
centro of the motion of b and c for the instant, and would corre- 
spond to the transmission of a different motion. The essential 
condition to be fulfilled by tooth outlines, in order that a con- 
stant velocity ratio may be maintained, may therefore be stated 
as follows: The tooth outlines must be such that their normal at 
the point oj contact shall always pass through the centro corre- 
sponding to the required velocity ratio. 

Fig. 199. 




Fig. 200. 



199. Given Tooth Outline to Find Form of Engaging Tooth. 

— Having given any curve that will serve for a tooth outline 

in one gear, the corresponding curve may be found in the other 

gear, which will engage with the given curve and transmit a 

m 
constant velocity ratio. Let — be the given velocity ratio. 

w + w = the sum of the radii of the two gears. Draw the line 
of centers AB, Fig. 201. Let P be the " pitch point," i.e., the 
point of contact of the pitch circles or the centro of relative 
motion of the two gears. To the right from P lay off a distance 
PB=m; from P toward the left lay off PA =n. A and B will 
then be the required centers of the wheels, and the pitch circles 



352 



MACHINE DESIGN. 



may be drawn through P. Let abc be any given curve on the 
wheel A. It is required to find the curve in B which shall engage 
with abc to transmit the constant velocity ratio required. A 
normal to the point of contact must pass through the centre 
If, therefore, any point, as a, be taken in the given curve, and a 
normal to the curve at that point be drawn, as aa, then when a 
is the point of contact, a will coincide with P. Also, if cy is a 
normal to the curve at c, then j will coincide with P when c is 
the point of contact between the gears; and since b is in the pitch 
line, it will itself coincide with P when it is the point of contact. 

Fig. 201. 




Fig. 202. 

Since the two pitch circles must roll upon each other without 
slipping, it follows that the arc Pa' = arc Pa, arc Pb f =arc Pb, 
and arc Pf = arc Py. 

Rotate the point a, about A, through the angle 6. At the 
same time a' rotates backward about B through the angle d r 
and a and a! coincide at P. Pa" represents the rotated position 
of the normal aa. Rotate Pa" about B through the angle f \ 
P will coincide with a' and a" will locate the point a' of the desired 
tooth outline of gear B. The point V of the desired outline is 
readily located by merely laying off arc Pb f =arc Pb. 

c f is located by the same method we employed to determine 
a! . This will give three points in the required curve, and through 



TOOTHED WHEELS OR GEARS. 353 

these the curve may be drawn. The curve could, of course, be 
more accurately determined by using more points. 

Many curves could be drawn that would not serve for tooth 
outlines; but, given any curve that will serve, the corresponding 
curve may be found. There would be, therefore, almost an 
infinite number of curves that would fulfill the requirements of 
correct tooth outlines. But in practice two kinds of curves are 
found so convenient that they are most commonly, though not 
exclusively, used. They are cycloidal and involute curves. 

200. Cycloidal Tooth Outlines. — It is assumed that the char- 
acter of cycloidal curves and method of drawing them is under- 
stood. 

In Fig. 202, let b and c be the pitch circles of a pair of wheels, 
always in contact at be. Also, let m be the describing circle in 
contact with both at the same point. M is the describing point. 
When one curve rolls upon another, the centro of their relative 
motion is always their point of contact. For, since the motion 
of rolling excludes slipping, the two bodies must be stationary, 
relative to each other, at their point of contact; and bodies that 
move relative to each other can have but one such stationary 
point in common — their centro. When, therefore, m rolls in 
or upon b or c, its centro relatively to either is their point of con- 
tact. The point M, therefore, must describe curves whose 
direction at any point is at right angles to a line joining that 
point to the point of contact of m with the pitch circles. Suppose 
the two circles b and e to revolve about their centers, being always 
in contact at be; suppose m to rotate at the same time about its 
center, the three circles being always in contact at one point and 
having no slip. The point M will then describe simultaneously 
a curve, b', on the plane of b, and a curve, c r , on the plane of c. 
Since M describes the curves simultaneously, it will always be 
the point of contact between them in any position. And since 
the point M moves always at right angles to a line which joins it 



354 MACHINE DESIGN. 

to be, therefore the normal to the tooth surfaces at their point 
of contact will always pass through be, and the condition for 
constant velocity ratio transmission is fulfilled. But these curves 
are precisely the epicycloid and hypocycloid that would be drawn 
by the point M in the generating circle, by rolling on the out- 
side of b and inside of c. Obviously, then, the epicycloids and 
hypocycloids generated in this way, used as tooth profiles, will 
transmit a constant velocity ratio. 

This proof is independent of the size of the generating circle, 
and its diameter may therefore equal the radius of c. Then the 
hypocycloids generated by rolling within c would be straight 
lines coinciding with the radius of c. In this case the flanks 
of the teeth of c become radial lines, and therefore the teeth are 
thinner at the base than at the pitch line; for this reason they 
are weaker than if a smaller generating circle had been used. All 
tooth curves generated with the same generating circle will work 
together, the pitch being the same. It is therefore necessary 
to use the same generating circle for a set of gears which need 
to interchange* 

The describing circle may be made still larger. In the first 
case the curves described have their convexity in the same direction, 
i.e., they lie on the same side of a common tangent. When the 
diameter of the describing circle is made equal to the radius 
of c, one curve becomes a straight-line tangent to the other curve. 
As the describing circle becomes still larger, the curves have their 
convexity in opposite directions. As the circle approximates 
equality with c, the hypocycloid in c grows shorter, and finally 
when the describing circle equals c, it becomes a point which is 
the generating point in c, which is now the generating circle. If 
this point could be replaced by a pin having no sensible diameter, 
it would engage with the epicycloid generated by it in the other 
gear to transmit a constant velocity ratio. But a pin without 

* See § 205. 




TOOTHED WHEELS OR GEARS. 355 

sensible diameter will not serve as a wheel-tooth, and a proper 
diameter must be assumed, and a new curve laid off to engage 
with it in the other gear. In Fig. 203, AB is the epicycloid gen- . 
ated by a point in the circumference of the 
other pitch circle. CD is the new curve 
drawn tangent to a series of positions of the 
]in as shown. The pin will engage with 
this curve, CD, and transmit the constant 
velocity ratio as required. In Fig. 202, let it FlG< 

be supposed that when the three circles rotate 
constantly tangent to each other at the pitch point be, a pencil is 
fastened at the point M in the circumference of the describing 
circle. If this pencil be supposed to mark simultaneously upo.i 
the planes of b, c, and that of the paper, it will describe upon b 
an epicycloid, on c a hypocycloid, and on the plane of the 
paper an arc of the describing circle. Since M is always 
the point of contact of the cycloidal curves (because it gen- 
erates them simultaneously), therefore, in cycloidal gear-teeth, 
the locus or path 0) the point of contact is an arc of the describ- 
ing circle. The ends of this path in any given case are located 
by the points at which the addendum circles cut the describing 
circles. 

In the cases already considered, where an epicycloid in one 
wheel engages with a hypocycloid in the other, the contact of 
the teeth with each other is all on one side of the line of centers. 
Thus, in Fig. 202, if the motion be reversed, the curves will be 
in contact until M returns to be along the arc MD-bc; but after 
M passes be contact will cease. If c were the driving-wheel, the 
point of contact would approach the line of centers; if b were the 
driving-wheel the point of contact would recede from the line of 
centers. Experience shows that the latter gives smoother running 
because of better conditions as regards friction between the tooth 
surfaces. It would be desirable, therefore, that the wheel with 
the epicycloid al curves should always be the driver. But it 



S3 6 MACHINE DESIGN. 

should be possible to use either wheel as driver to meet the varying 
conditions in practice. 

Another reason why contact should not be all on one side 
of the line of centers may be explained as follows : 

201. Definitions: Pitch-arc, Arc of Action, Line of Pressure. 
— The angle through which a gear-wheel turns while one of its 
teeth is in contact with the corresponding tooth in the other gear 
is called the angle of action. It is found by drawing radial lines 
from the center to the pitch circle at the two tooth positions cor- 
responding to the beginning and end of engagement. The arc of 
the pitch circle corresponding to the angle of action is called the 
arc of action. 

The arc of action must be greater than the " pitch arc " (the arc 
of the pitch circle that includes one tooth and one space), or else 
contact will cease between one pair of teeth before it begins between 
the next pair. Constrainment would therefore not be complete, 
and irregular velocity ratio with noisy action would result. 

In Fig. 204, let AB and CD be the pitch circles of a pair of 
gears and E the describing circle. Let an arc of action be laid 
off on each of the circles from P, as Pa, Pc, and Pe. Through 
e, about the center O, draw an addendum circle; i.e., the circle 
which limits the points of the teeth. Since the circle E is the 
path of the point of contact, and since the addendum circle 
limits the points of the teeth, their intersection, e, is the point at 
which contact ceases, rotation being as indicated by the arrow. 
If the pitch arc just equals the assumed arc of action, contact 
will be just beginning at P when it ceases at e; but if the pitch 
arc be greater than the arc of action, contact will not begin at 
P till after it has ceased at e, and there will be an interval when 
AB will not drive CD. The greater the arc of action the greater 
the distance of e from P on the circumference of the describing 
circle. The direction of pressure between the teeth is always 
a normal to the tooth surface, and this always passes through 
the pitch point; therefore the greater the arc of action — i.e., 



TOOTHED WHEELS OR GEARS. 



357 



the greater the distance of e from P— the greater the obliquity 
of the line of pressure. The pressure may be resolved into two 
components, one at right angles to the line of centers and the 
other parallel to it. The first is resisted by the teeth of the follower- 
wheel, and is effective to produce the desired rotation, while the 
second tends to crowd the journals apart, and therefore produces 
pressure with resulting friction. Hence it follows that the greater 
the arc of action the greater will be the average obliquity of 
the line of pressure, and therefore the greater the component 





Fig. 204. 



Fig. 205. 



of the pressure that produces wasteful friction. If it can be 
arranged so that the arc of action shall be partly on each 
side of the line of centers, the arc of action may be made 
greater than the pitch arc (usually equal to about i\ times 
the pitch arc) ; then the obliquity of the pressure-line may 
be kept within reasonable limits, contact between the teeth 
will be insured in all positions, and either wheel may be the 
driver. This is accomplished by using two describing circles as 
in Fig. 205. Suppose the four circles A, B, a, and j3 to roll con- 
stantly tangent at P. a will describe an epicycloid on the plane 
of B, and a hypocycloid on the plane of A. These curves will 
engage with each other to drive correctly. /? will describe an 
epicycloid on A, and a hypocycloid on B. These curves will 
engage, also, to drive correctly. If the epicycloid and hypocycloid 
in each gear be drawn through the same point on the pitch circle, 
a double curve tooth outline will be located, and one curve will 



358 



MACHINE DESIGN. 



engage on one side of the line of centers and the other on the 
other side. If A drives as indicated by the arrow, contact will 
begin at D, and the point of contact will follow an arc of a to P, 
and then an arc of /? to C 

202. Involute Tooth Outlines. — If a string is wound around a 
cylinder and a pencil-point attached to its end, thi s point will 
trace an involute on a plane normal to the axis of the cylinder 
as the string is unwound from the cylinder. Or, if the point 
be constrained to follow a tangent to the cylinder, and the string 
be unwound by rotating the cylinder about its axis, £he point 
will trace an involute on a plane that rotates with the cylinder. 

Illustration. — Let a, Fig. 206, be a circular piece of wood free 
to rotate about C; /? is a circular piece of cardboard made fast 
to a; AB is a straight-edge held on the circumference of a f 
having a pencil-point at B. As B moves along the straight-edge 
to A, a and /? rotate about C, and B traces an involute DB upon 
/?, the relative motion of the tracing point and /? being just the 
same as if the string had been simply unwound from a fixed* 
If the tracing point is caused to return along the straight-edge 
it will trace the involute BD in a reverse direction. 





Fig. 206. Fig. 207. 

The centro of the tracing point is always the point of tan- 
gency of the string with the cylinder; therefore the string, or 
straight-edge, in Fig. 208, is always at right angles to the direc- 
tion of motion of the tracing point, and hence is always a normal 
to the involute curve. Let a and 8, Fig. 207, be two base cylin- 
ders ; let AB be a cord wound upon a and /? and passing through 
the centro P, which corresponds to the required velocity ratio. 



TOOTHED WHEELS OR GEARS. 359 

Let a and /? be supposed to rotate so that the cord is wound from 
P upon a. Then any point in the cord will move from A toward 
B, and, if it be a tracing-point, will trace an involute of a on the 
plane of a (extended beyond the base cylinder) , and will also trace 
an involute of /? upon the plane of /?. These two involutes will 
serve for tooth profiles for the transmission of the required con- 
stant velocity ratio, because AB is the constant normal to both 
curves at their point of contact, and it passes through P, the 
centro corresponding to the required velocity ratio. Hence the 
necessary condition is fulfilled. The pitch circles will have OP 
and O'P as their respective radii. 

Since a point in the line AB describes the two involute curves 
simultaneously, the point of contact of the curves is always in 
the line AB. And hence AB is the path of the point of contact. 
In any given case the two ends of the path he at the intersections 
of the addendum circles with AB. Should either addendum 
circle intersect the line of action outside of the portion lying 
between A and B, " interference " takes place. In such cases 
the addendum may be shortened or the profile of the tooth- 
point modified from the true involute. 

To avoid interference, the allowable length of addendum 
in any case (conversely, the least number of teeth of pinion) 
may be computed by Bach's formula: 

NJ . (N, 
a = 



27T \2N_ 

a = addendum ( commonly = - = - ) , inches; 

N\ = number of teeth of pinion; 
N 2 = number of teeth of gear (= 00 for rack); 
a = angle between line of action and line of centers ; 
P = circular pitch, inches. 
One of the advantages of involute curves for tooth profiles 
is that a change in distance between centers of the gears docs 
not interfere with the transmission of a constant velocity ratio. 



360 MACHINE DESIGN. 

»e proved as follows : In Fig. 207, 

; that is, the ratio of the radii of the base circles (i.e., 



This may be proved as follows : In Fig. 207, from similar triangles 
OB OP 



O f A O'P 

sections of the base cylinders) is equal to the ratio of the radii of 
the pitch circles. This ratio equals the inverse ratio of angular 
velocities of the gears. Suppose now that O and O r be moved 
nearer together; the pitch circles will be smaller, but the ratio 
of their radii must be equal to the unchanged ratio of the 
radii of the base circles, and therefore the velocity ratio remains 
unchanged. Also the involute curves, since they are generated 
from the same base cylinders, will be the same as before, and 
therefore, with the same tooth outlines, the same constant velocity 
ratio will be transmitted as before. 

203. Racks. — A rack is a wheel whose pitch radius is infinite; 
its pitch circle, therefore, becomes a straight line, and is tangent 
to the pitch circle of the wheel, or pinion,* with which the rack 
engages. The line of centers is a normal to the pitch line of the 
rack, through the center of the pitch circle of the pinion. The 
pitch of the rack is determined by laying off the circular pitch 
of the engaging wheel on the pitch line of the rack. The curves 
of the cycloidal rack-teeth, like those of wheels of finite radius, 
may be generated by a point in the circumference of a circle which 
rolls on the pitch circle. Since, however, the pitch circle is now 
a straight line, the tooth curves will be cycloids, both for flanks 
and faces. In Fig. 208, AB is the pitch circle of the pinion and 
CD is the pitch line of the rack; a and b are describing circles. 
Suppose, as before, that all move without slipping and are con- 
stantly tangent at P. A point in the circumference of a will then 
describe simultaneously a cycloid on CD, and a hypocycloid 
within AB. These will be correct tooth outlines. Also, a point 
in the circumference of b will describe a cycloid on CD, and an 
epicycloid on AB. These will be correct tooth outlines. To 



* Pinion is a word to denote a gear having a low number of teeth, or the smaller 
one of a pair of engaging gears. 



TOOTHED WHEELS OR GEARS. 361 

find the path of the point of contact, draw the addendum circle 
EF of the pinion, and the addendum line GH of the rack. When 
the pinion turns clockwise and drives 
the rack, contact will begin at e and 

-c 





Fig. 208. 

follow arcs of the describing circles through P to K. It is ob- 
vious that a rack cannot be used where rotation is continuous in 
one direction, but only where motion is reversed. 

Involute curves may also be used for the outlines of rack 
teeth. Let CD and CD', Fig. 209, be the pitch lines. When it 
is required to generate involute curves for tooth outlines, for a 
pair of gears of finite radius, a line is drawn through the pitch 
point at a given angle to the line of centers (usually 75 ) ; this 
line is the path of the point which generates two involutes simul- 
taneously, and therefore the path of the point of contact between 
the tooth curves. It is also the common tangent to the two base 
circles, which may now be drawn and used for the describing of 
the involutes. To apply this to the case of a rack and pinion, 
draw EF, Fig. 209, making the desired angle with the line of 
centers, OP. The base circles must be drawn tangent to this 
line; AB will therefore be the base circle for the pinion. But 
the base circle in the rack has an infinite radius, and a circle of 
infinite radius drawn tangent to EF would be a straight line 
coincident with EF. Therefore EF is the base line of the rack. 
But an involute to a base circle of infinite radius is a straight 
line normal to the circumference — in this case a straight line per- 
pendicular to EF. Therefore the tooth profiles of a rack in the 
involute system will always be straight lines perpendicular to the 
path of the describing point, and passing through the pitch points. 
If, in Fig. 209, the pinion move clockwise and drive the rack, the 



362 



MACHINE DESIGN. 



contact will begin at E f the intersection of the addendum line 
of the rack GH, and the path of the point of contact EF, and 
will follow the line EF through P to the point where EF cuts the 
addendum circle LM of the pinion. 

204. Annular Gears. — Both centers of a pair of gears may be 
on the same side of the pitch point. This arrangement corre- 
sponds to what is known as an annular gear and pinion. Thus, 
in Fig. 210, AB and CD are the pitch circles, and their centers 
are both above the pitch point P. 
Teeth may be constructed to trans- ' 





Fig. 210. 



Fig. 211. 



mit rotation between AB and CD. AB will be an ordinary 
spur pinion, but it is obvious that CD becomes a ring of meta! 
with teeth on the inside, i.e., it is an annular gear. In this case 
a and /? may be describing circles for cycloidal teeth, and a point 
in the circumference of a will describe hypocycloids simultaneously 
on the planes of AB and CD; and a point in the circumference 
of /? will describe epicycloids simultaneously on the planes of AB 
and CD. These will engage to transmit a constant velocity 
ratio. Obviously the space inside of an annular gear corresponds 
to a spur-gear of the same pitch and pitch diameter, with tooth 
curves drawn with the same describing circle. Let EF and GH, 
Fig. 210, be the addendum circles. If the pinion move clockwise 
driving the annular gear, the path of the point of contact will be 
from e along the circumference of a to P, and from P along the 
circumference of /? to K. 

The construction of involute teeth for an annular gear and 
pinion involves exactly the same principles as in the case of a 



TOOTHED IV HEELS OR GE4RS. 363 

pair of spur-gears. The only difference of detail is that the 
describing point is in the tangent to the base circles produced 
instead of being between the points of tangency. Let O and O' ', 
Fig. 211, be the centers, and AB and I J the pitch circles of an 
annular gear and pinion. Through P, the point of tangency of 
the pitch circles, draw the path of the point of contact at the given 
angle with the line of centers. With O and O' as centers draw 
tangent circles to this line. These will be the involute base 
circles. Let the tangent be replaced by a cord, made fast, say, 
at K', winding on the circumference of the base circle CK! , to D, 
and then around the base circle FE in the direction of the arrow, 
and passing over the pulley G, which holds it in line with PB. 
If rotation be supposed to occur with the two pitch circles always 
tangent at P without slipping, any point in the cord beyond P 
toward G will describe an involute on the plane //, and another 
on the plane of AB. These will be the correct involute tooth 
profiles required. Draw NQ and LM, the addendum circles. 
Then if the pinion move clockwise, driving the annular gear, 
the point of contact starts from e and moves along the line GH 
through P to K. 

When a pair of spur-gears mesh with each other, the direction 
of rotation is reversed. But an annular gear and pinion meshing 
together rotate in the same direction. 

205. Interchangeable Sets of Gears. — In practice it is desirable 
to have "interchangeable sets " of gears; i.e., sets in which any 
gear will "mesh " correctly with any other, from the smallest 
pinion to the rack, and in which, except for limiting conditions 
of size, any spur-gear will mesh with any annular gear. Inter- 
changeable sets may be made in either the cycloidal or involute 
system. A necessary condition in any set is that the pitch shall 
be constant, because the thickness of tooth on the pitch line must 
always equal the width of the space (less backlash). If this 
condition is unfulfilled they cannot engage, whatever the form 
of the tooth outlines. 



364 MACHINE DESIGN. 

The second condition for an interchangeable set in the cycloi- 
dal system is that the size of the describing circle shall be constant. 
If the diameter of the describing circle equals the radius of the 
smallest pinion's pitch circle, the flanks of this pinion's teeth will 
be radial lines, and the tooth will therefore be thinner at the base 
than at the pitch line. As the gears increase in size with this 
constant size of describing circle, the teeth grow thicker at the 
base; hence the weakest teeth are those of the smallest pinion. 

It is found unadvisable to make a pinion with less than twelve 
teeth. If the radius of a fifteen-tooth pinion be selected for the 
diameter of the describing circle, the flanks which bound a space 
in a twelve-tooth pinion will be very nearly parallel, and may 
therefore be cut with a milling-cutter. This would not be possi- 
ble if the describing circle were made larger, causing the space 
to become wider at the bottom than at the pitch circle. There- 
fore the maximum describing circle for milled gears is one whose 
diameter equals the pitch radius of a fifteen-tooth pinion and 
it is the one usually selected. Each change in the number of 
teeth with constant pitch causes a change in the size of the pitch 
circle. Hence the form of the tooth outline, generated by a 
describing circle of constant diameter, also changes. For any 
pitch, therefore, a separate cutter would be required corresponding 
to every number of teeth, to insure absolute accuracy. Practi- 
cally, however, this is not necessary. The change in the form 
of tooth outline is much greater in a small gear, for any increase 
in the number of teeth, than in a large one. It is found that 
twenty-four cutters will cut all possible gears of the same pitch 
with sufficient practical accuracy. The range of these cutters 
is indicated in the following table, taken from Brown and Sharpe's 
"Treatise on Gearing." 

These same principles of interchangeable sets of gears with cy- 
cloidal tooth outlines apply not only to small milled gears as above, 
but also to large cast gears with tooled or un tooled tooth surfaces. 



TOOTHED WHEELS OR GEARS 



365 









Table 


XXXI. 






liter A cuts 


12 


teeth 




Cutter M cuts 27 to 29 teet 


" B " 


13 


" 




1 


N " 


30 " 33 


,< C «« 


14 


< t 




" 


" 


34 " 37 " 


it D « 


IS 


< i 




< c 


p « 


38 " 42 " 


" E " 


16 


< i 




( c 


Q " 


43 " 49 


<< F > 


1 7 


" 




" 


R « 


5° " 59 


" G " 


18 


( 1 




" 


S « 


60 " 74 


" H " 


19 


1 < 




( < 


T ' 


75 " 99 


" I " 


20 


" 




" 


U ' 


100 " 149 " 


" J " 


21 


to 22 


teeth 


11 


V < 


150 " 249 " 


" K " 


23 


" 24 


( < 


" 


W ' 


250 " rack 


" L " 


25 


" 26 


< < 


" 


X ' 


rack 



206. Interchangeable Involute Gears. — In the involute system 
the second condition of interchangeability is that the angle between 
the common tangent to the base circles and the line of centers shall 
be constant. This may be shown as follows: Draw the line of 
centers, AB, Fig. 212. Through P, the assumed pitch point, 
draw CD, and let it be the constant common tangent to all base 




circles from which involute tooth curves are to be drawn. Draw 
any pair of pitch circles tangent at P, with their centers in the 
line AB. About these centers draw circles tangent to CD; these 
are base circles, and CD may represent a cord that winds from 
one upon the other. A point in this cord will generate simul- 
taneously involutes that will engage for the transmission of a con- 
stant velocity ratio. But this is true of any pair of circles that 
have their centers in AB, and are tangent to CD. Therefore, 



;66 



MACHINE DESIGN. 



if the pitch is constant, any pair of gears that have the base circles 
tangent to the line CD will mesh together properly. As in the 
cycloidal gears, the involute tooth curves vary with a variation 
in the number of teeth, and, for absolute theoretical accuracy, 
there would be required for each pitch as many cutters as there 
are gears with different numbers of teeth. The variation is 
least at the pitch line, and increases with the distance from it. 
The involute teeth are usually used for the finer pitches and the 
cycloidal teeth for the coarser pitches ; and since the amount that 
the tooth surface extends beyond the pitch line increases with the 
pitch, it follows that the variation in form of tooth curves is 
greater in the coarse pitch cycloidal gears than in the fine pitch 
involute gears. For this reason, with involute gears, it is only 
necessary to use eight cutters for each pitch. The range is shown 
in the following table, which is also taken from Brown and Sharpe's 
"Treatise on Gearing " : 

Table XXXII. 
No. i will cut wheels from 135 teeth to racks 



<< 2 a 


" 


"■ 55 " ' 


J 34 


inclusive 


c< 3 << 


" 


" 35 " ' 


54 


'* 


" 4 " 


< 1 


" 26 " 


34 


" 


" 5 " 


1 < 


" 21 ' ' ' 


25 


' ' 


" 6 " 


' 


« 17 << < 


20 




" 7 " 


" 


" 14 " « 


16 


< t 


" 8 " 


C i 


12 ' < 


13 


li 



207. Laying Out Gear-teeth. — Exact and Approximate 
Methods. — There is ordinarily no reason why an exact method 
for laying out cycloidal or involute curves for tooth outlines should 
not be used, either for large gears or gear patterns, or in making 
drawings. It is required to lay out a cycloidal gear. The pitch, 
and diameters of pitch circle and describing circle are given. 

Draw the pitch circle on the drawing-paper, using a fine line. 
On a flat piece of tracing- cloth or thin, transparent celluloid draw 
a circle the size of the generating circle. Use a fine, clear line. 



TOOTHED WHEELS OR GEARS 



367 



Place it over the drawn pitch circle so that it is tangent to the 
latter at P as shown in Fig. 213. AB is an arc of the pitch circle. 
Insert a needle point at P, and using it as a pivot swing the 
tracing-cloth in the direction of the arrow a very short distance, 
so that the generating circle cuts the pitch circle at a new point 
Q, as shown exaggeratedly in Fig. 214. Q should be taken very close 



Epicycloid 





" Fig. 213. Fig. 214. Fig. 215. Fig. 216. 

to P. Insert a needle point at Q, remove the one at P, and swing 
the cloth, about Q as a pivot, in the direction of the arrow until 
the two circles are tangent at Q. (See Fig. 215.) The point P 
of the tracing-cloth now lies off the pitch circle a short distance. 
With a needle point prick its present position through to the 
drawing-paper. Now with Q as a pivot rotate the tracing-cloth 
until the two circles intersect at a point R slightly beyond Q. 
Insert needle at R and remove the one at Q. Swing tracing- 
cloth about R until R becomes the point of tangency of the two 
circles and then prick the new position of P through to the drawing- 
paper. Taking points very near together and repeating the 
operation gives a close approximation to true rolling of the gen- 
erating circle on the pitch circle and therefore the path of the 
point P as marked on the drawing-paper by pricked points is an 
epicycloid and may be used for the face of the tooth. 

Next place the tracing-cloth on the inside of the pitch circle, 
as shown in Fig. 216, with the generating circle tangent to the 
pitch circle at the original point P. Using the method just 
described to prevent slipping, roll the generating circle, in the 
direction of the arrow, on the pitch circle and the path traced 
by P as marked by pricked points on the drawing-paper will be 
a hypocycloid for the flank of the tooth. 



3 68 



MACHINE DESIGN. 



The compound curve aPb, Fig. 217, has now been traced, 
which forms the basis of the completed tooth outline. 

AB is an arc of the pitch circle whose center is at O. With 
O as center, swing in the addendum circle CD and the full 
depth circle EF, according to the proportions given in § 208. 
With a radius equal to T * F of the circular pitch draw the fillet 
cd tangent to EF and aPb. The completed tooth profile is the 
curve cdPe. 

Cut a wooden template to fit the tooth curve, and make it 
fast to a wooden arm free to rotate about O, making the edge of 
the template coincide with cdPe. It may now be swung succes- 
sively to the other pitch points, and the tooth outline may be drawn 
by the template edge. This gives one side of all of the teeth. 
The arm may now be turned over and the other sides of the 
teeth may be drawn similarly. 





Fig. 2i* 



It is required to lay out exact involute teeth. The pitch, 
pitch- circle diameter, and angle of the common tangent are 
given. — Draw the pitch circle AB, Fig. 218, and the line of 
centers 00' . Through the pitch point P draw CD, the common 
tangent to the base circles, making the angle /? with the line of 
centers. Draw the base circle EF about O tangent to CD. 

On a piece of flat tracing-cloth draw a fine, clear, straight line 
and lay the tracing-cloth over the drawing so that this line coin- 
cides with CD. Take a needle point and insert it at the point 



TOOTHED W HEELS OR GEARS. 369 

of tangency Q. With another needle, mark the point P on the 
tracing-cloth. Now, employing a pair of needle points to pre- 
vent slipping, roll the traced line on the base circle EF, prick- 
ing the path, aPb, of the point P of the tracing-cloth through to 
the drawing-paper. This path is the involute of the base circle 
and is the basis of the involute tooth outline. To complete the 
latter proceed as follows: Draw the addendum circle GH and 
the full-depth circle JK. In general JK will lie inside of the 
base circle EF and it will be necessary to extend the tooth outline 
inward beyond a. About O swing a circle whose diameter equals 
one hall the circular pitch and draw ac tangent to it. With a 
radius equal to T V of the circular pitch swing in the fillet de tan- 
gent to JK and ac. deaPj is the completed outline. If the 
gear has 20 teeth or less ac should be made a radial line. If 
EF lies inside of JK we draw the fillet tangent to JK and aP.* 

208. Gear Proportions. — The following formulas and table 
are given to assist in the practical proportioning of gears : 
Let D = pitch diameter; 
D\ = outside diameter; 

D 2 = diameter of a circle through the bottom of spaces; 
P = circular pitch = space on the pitch circle occupied by 

a tooth and a space ; 
p = diametral pitch = number of teeth per inch of pitch- 
circle diameter; 
N =numbcr of teeth; 
t = thickness of tooth on pitch line; 
a = addendum; 

* Approximate tooth outlines may be drawn by the use of instruments, such 
as the Willis odontograph, which locates the centers of approximate circular arcs; 
the templet odontograph, invented by Prof. S. W. Robinson; or by some geomet- 
rical or tabular method for the location of the centers of approximate circular 
arcs. For descriptions, see "Elements of Mechanism," Willis; "Kinematics," 
McCord; "Teeth of Gears," George B. Grant; "Treatise on Gearing," pub- 
lished by Brown and Sharpe. 



37o 



MACHINE DESIGN. 



c = clearance; 

d = working depth of spaces; 

d\ = full depth of spaces. 



Then 



N + 2 





v 1 = 


P 


1 


D 2 = 


D- 2 (a+c); 




N = 


Dtz 

p ' 


P = 


Dtz 


D 


PN 

TZ ' 


N = 


Dp; 




p= 


N 
"D ' 


D 


N 
~ P 


P P = 


■-*', 




TZ 


; p 


TZ 


t = 


P 

2 


TZ 

~7' n0 
2p 


backlash ; 


P 

20 


TZ 
p2Q 


; d 


= 2d 


; di 


= 2d + 


c; a 


=— inches 
P 



10 

The following dimensions are given as a guide; they may be 
varied as conditions of design require: Width of face = about 3P; 
thickness of rim =1.25/; thickness of arms = 1.25/, no taper. 
The rim may be reinforced by a rib, as shown in Fig. 219. Diam- 
eter of hub = 2 X diameter of shaft. Length of hub = width of 
face + J"; width of arm at junction with hub=J circumference 
of the hub for six arms. Make arms taper about J" per foot on 
each side. 

Table XXXIII. 



Diametral 
Pitch. 


Circular Pitch. 


Thickness of 
Tooth on the 


Diametral /-•:»«.. 
Pitch. Circu 


lar Pitch. 


Thickness of 
Tooth on the 




Pitch Line. 




Pitch Line. 


P 


P 


i 


P 


P 


t 


i 


6.283 


3-I4I 


3* 


897 


•449 


I 


4 


189 


2 


094 


4 


785 




393 


1 


3 


141 


1 


57i 


5 


628 




314 


ii 


2 


5.13 


1 


256 


6 


5 2 3 




262 


i* 


2 


094 


1 


047 


7 


449 




224 


if 


1 


795 




897 


8 


393 




196 


2 


1 


57i 




785 


9 


349 




174 


a* 


1 


39 6 




698 


10 


3i4 




157 


a* 


1 


256 




628 


12 


262 




131 


a* 


1 


142 




57i 


14 


224 




112 


3 


1.047 


•5 2 3 









TOOTHED WHEELS OR GE4RS. 371 

209. Strength of Gear-teeth. — The maximum work trans- 
mitted by a shaft per unit time may usually be accurately estimated ; 
and, if the rate of rotation is known, the torsional moment may 
be found. Let O, Fig. 220, represent the axis of a shaft perpen- 
dicular to the paper. Let A = maximum work to be transmitted 
per minute; let N = revolutions per minute; let Fr = torsional 
moment. Then F is the force factor of the work transmitted, 
and 2rcrN is the space factor of the work transmitted. Hence 

2FnrN =A, and Fr = torsional moment =■ — =r=. 

27ZN 

If the work is to be transmitted to another shaft by means of 
a spur-gear whose radius is ri, then for equilibrium F\r x =Fr, 

Fr 

and F\ = — . -Fi is the force at the pitch surface of the gear 

whose radius is r x , i.e., it is the force to be sustained by the gear- 
teeth. Hence, in general, the force sustained by the teeth 0] a gear 
equals the torsional moment divided by the pitch radius 0] the gear. 

When the maximum force to be sustained is known the teeth 
may be given proper proportions. The dimensions upon which 
the tooth depends for strength are : Thickness of tooth = /, width 
of face of gear = 5, and depth of space between teeth =/. These 
all become known when the pitch is known, because / is fixed for 
any pitch, and / and b have values dictated by good practice. 
The value of b may be varied through quite a range to meet the 
requirements of any special case. 

In computations for strength the tooth is treated as a canti- 
lever. It has been customary to consider that the entire load 
is borne by a single tooth (i.e., that there is contact between only 
one pair of teeth), the point of application being the extreme 
end of the tooth, and the direction of the acting force being 
normal to the tooth profile at that point. This assumes that 
the arc of action is no greater than the pitch arc, which may 
be true of a pair of gears having a low number of teeth; but 



372 



MACHINE DESIGN. 



in all cases in which the arc of action exceeds the pitch arc the 
force is borne by several pairs of teeth in simultaneous engage- 
ment, and to consider it borne by a single pair leads to an excess 
of strength. This is of course an assumption on the safe side. 
Experimentally determined coefficients to correct for the ratio of 
the arc of action to the pitch arc will be found in Table XXXVIII. 
It is also assumed that the load is uniformly distributed across 
the face of the tooth. This is a safe assumption if the width of 
face, b, does not exceed three times the circular pitch, i.e., 3P, and 
if the gears are well aligned and rigidly supported. All teeth of 
the same pitch have not the same form, as was explained in the 
discussion of interchangeable gears, and therefore they vary in 
strength. The fewer teeth the thinner they will be at their base 
and consequently the weaker they will be when acting as canti- 
levers. 






Fig. 219. 



Fig. 220. 



Fig 221. 



Mr. Wilfred Lewis * has drawn a number of figures on a 
large scale to represent very accurately the teeth cut by complete 
s^ts of cutters of the 15 involute, the 20 involute, and the cycloidal 
systems. In the latter he used a rolling circle having a diameter 
equal to the radius of the 12-tooth pinion. The proportions 
of the teeth used in his investigation are slightly different from 
those given above which correspond to the Brown and Sharpe 
system, but no serious errors will result from applying the formulas 
derived by him. His reasoning was as follows (see Fig. 221): 



* Proc. Phila. Eng. Club, 1893, and Amer Mach., May 4 and June 22, 1893. 



TOOTHED IV HEELS OR GEARS, 373 

The greatest stress in the tooth occurs when the load is applied 
at the end of the tooth as indicated by the arrow at a, its direction 
of action being normal to the profile at a. The component of 
this force, which, is effective to produce rotation of the gear, equals 
F and is called the working force. 

This load is applied at b and induces a transverse stress in 
the tooth. To determine where the tooth is weakest advantage 
is taken of the fact that any parabola in the axis be and tangent to 
bF incloses a beam of uniform strength. Of all the parabolas 
that may thus be drawn one only will be tangent to the tooth form 
(as shown by the dotted line in the figure) and the weakest sec- 
tion of the tooth will be that through the points of tangency c 
and d. Having determined the weakest section in each case, Mr. 
Lewis developed the following general formulae from the data so 
obtained : 

For 1 5 involute and the cycloidal system, using a rolling 
circle whose diameter equals the radius of the 12-tooth pinion, 

F=jPb\o.i24-—jy-U 



For the 20 involute system 

i ? =/P & (o.x54-°-^).* 



F= working force in pounds; 

/=safe allowable unit stress in pounds per 1 square inch; 
P= circular pitch in inches; 
b = width of face of gear in inches; 
iV=number of teeth in the gear. 



* For the cycloidal system, using a rolling circle whose diameter equals the 
radius of the 15-tooth pinion, 



*/»(«*-2j2). 

(Trans. A. S. M. E., Vol. XVIII, p. 776.) 



374 



MACHINE DESIGN. 



Experimental data fixing the value of / for different materials 
and velocities have been lacking. 

Experiments made at Stanford University * lead to the fol- 
lowing equation for 14J standard, involute, cast iron, cut gears : 



F = 



f»Pb_ 
K 



(». 



154- 



1.26 

~N 



VOL. 



F = sa,ie working load at pitch line, in pounds; 

f b = modulus of rupture in flexure 

= 36,000 lbs. per square inch for cast iron; 

P = circular pitch, inches ; 
b = width of face, inches; 

K = factor of safety 

= 4 for uniform stress in one direction only 

= 6 for suddenly applied loads, one direction only 

= 8 for shocks and reversals of stress ; 

v = velocity coefficient from Table XXXIV 

a = arc of action coefficient from Table XXXV; 

N = number of teeth in gear. 



Table XXXIV. — Velocity Coefficients, 14I Involute Gears. 



Pitch velocity in feet 

per minute 

Coefficient, v 



1. 00 


100 
0.80 


200 
o-73 


30c 

0.68 


400 
0.64 


500 
O.60 


750 

o-55 


1000 
0.50 


1250 
o.45 


1500 
0.43 


1750 
0.41 


2000 
0.40 


Table XXXV. — Arc of Action Coefficients, 145 ° Involute Gears. 


_ . Arc of action 
Ratio: . , 

Pitch arc 

Coefficient, a 


1 

1 


1.4 
1.05 


1.6 

1 .10 


1.7 1.8 

I. 15 1 I. 24 


1.9 
i.*8 


i-95 
1.47 


2.00 
1 .60 


2.20 
1 .60 

































* The Strength of Gear Teeth, G. H. Marx, Trans. A. S. M. E., 191 2; and 
G. H. Marx and L. E. Cutter, Trans. A.S.M.E, 1915. 



TOOTHED IVHEELS OR GE/1RS. 



375 



For 2o° involute, stub-tooth, cast-iron, cut gears the same 
investigation yielded the formula: 

_ f b Pb( 2.6o\ 

The values of v and a to be used in this equation are given in 
Tables XXXVI and XXXVII. 



Table XXXVI. — Velocity Coefficients, 20 Involute, 


Stub 


-tooth Gears. 


Pitch vel., ft. per 
min 





100 


200 


300 


400 


500 


750 


1000 


1250 


1500 
0.50 


1750 

0.47 








Coefficient, v 


1 .00 


0.83 


0.76 


0.71 


0.67 


0.64 


o-59 


o-55 


0.52 


0.45 



Table XXXVII. — Arc of Action Coefficients, 20 Involute, Stub-tooth 

Gears. 



_. . Arc of action 

Ratio : - — =^ — : 

Pitch arc 


1 .00 


1.23 


1-37 


i-43 


1.46 


1.47 


1.49 


i-53 


1.56 


1. 61 


Coefficient, a. . . 


1 .00 


1 -13 


1 .20 


1.24 


1. 25 


1 . 26 


1 . 27 


1 . 29 


131 


i-33 





To save computation of the arc of action, Table XXXVIII 
gives the values of a for both forms for typical sets of gears. 
Interpolations can be made readily for other combinations. 



Table XXXVIII.— Values of a. 



Teeth in Engaging Gears. 


Corresponding a. 






I4i° Invol. 


20 Invol. Stub 
Tooth. 


Single tooth engages 


1 .00 


I .OO 


12 


12 


1 . 10 


1 


13 


20 


30 


115 


1 


20 


30 


30 


i-47 


1 


22 


30 


40 


1 .60 


I 


24 


30 


60 


1 .60 


I 


25 


30 


80 


1 .60 


1 


26 


30 


100 


1.60 


I 


27 


30 


Rack 


1 .60 


1 


29 


100 


100 


1 .60 


I 


31 


100 


Rack 


1 .60 


I 


33 



376 MACHINE DESIGN. 

Experiments reported by Mr. Andrew Gleason * show the 
results of tests on 14 tooth steel pinions, 14J involute, i-inch 
face, of various kinds of steel, soft, case-hardened, and tempered. 
His results indicate that for soft, 0.20 carbon, open-hearth 
steel a value of f b of 60,000 will not exceed the elastic limit. The 
same material case hardened and heat treated showed an ultimate 
breaking value of f b in excess of 170,000. Nickel steel gave 
values of f b about 15 per cent higher than these; and chrome- 
nickel tempering steel values about 50 per cent higher. Limited 
in number though they are, these experiments are very significant. 

For rawhide gears, — may be taken as 5000 lbs. per square 

inch as a minimum. Hard fibre is more brittle; -^ = 5000 may 

be taken as a maximum. 

In these formulae, for a given gear the whole right side of 
the equation becomes known and the allowable value of F is 
readily determined. It is more difficult to apply the formulas 
where the force to be transmitted is given. In such a case the 
value of P is determined by trial. 

210. Problem. — Design a pair of 14^° involute gears to transmit 
6 H.P. The distance between centers is 10 inches and the 
velocity ratio of the shafts is to be f. The pinion shaft makes 
150 revolutions per minute. 

The distance between centers being 10 inches and the velocity 
ration §, the radii will be to each other as 3:2 and their sum 
= 10 inches, hence the radius of the pinion will be 4 inches, while 
that of the gear will be 6 inches. If both gears are of the same 
material the teeth of the smaller will be the weaker. Com- 
putations will therefore be made for the pinion because the gear 
will be stronger and, consequently, safe. The pitch diameter 

* Machinery, Jan., 19 14. 



TOOTHED IVHEELS OR GEARS. 377 

of the pinion =8 inches, its velocity = 150X71 X-f? =314.2 feet per 
minute. 

6 H.P. =6X33,000 = 198,000 ft. -lbs. per minute; 

198000 
F = — =632 lbs. 

Assuming cast iron, 14^° involute gears, variable load in 
one direction only: 

_ fJPbf I.26\ 

F = 6t ) 2 lbs. P is sought, b is unknown, but a trial value, 
proportionate to the size of the gear diameters, may be assumed; 
if the value assumed is too small it leads to an irrational quadratic 
equation for P; if too great a value is used P will be less than 
J6. In this case assume first trial value of b = 1 inch. N is 

unknown, so -— is substituted for it, D\ being the pitch diameter 

of the pinion, 8 inches in this case. The velocity being 314.2 

feet per minute, the value of the velocity coefficient, v, from Table 

XXXIV, is 0.68. The ratio of arc of action to pitch arc being 

unknown it is safe to take a = 1.00. f b = 36,000 lbs. per square 

inch. K = 6. 

36,oooXPXi/ I.26XP\ ^ 

632 = ^ .154 J0.68X1, 

6 V 2 5- I 3 / 

/. P 2 -3.o8P + 2. 3 7i6= -0.6184, 



:. P - 1.54 = ±v -0.6184. 

This shows that the assumed trial value of b=\ inch was 
too small. Try 6 = 2 inches. 

36,000 XPX 2/ i.26P\ ^ 

632 = ^ .154 J0.68X1. 

6 \ 25.13/ 



378 MACHINE DESIGN. 



.'. P-l.54=±Vo.8 3 . 
.*. P = 0.63 (smaller root). 

Nearest regular diametral pitch p = $. 
This gives 5 X 8 = 40 teeth on pinion, and 
5 X 12 = 60 teeth on gear. 

These values will answer, although the ratio of b to P is a 
little more than the maximum desirable value of 3. A new trial 
of &=ij inches would lead to a value of P of about 0.92. The 
nearest regular value of p would be 3J, giving the pinion and 
gear, 28 and 42 teeth, respectively. 

If the greatest allowable value of b still leaves the imaginary, 
then the value of f b must be increased either by using a stronger 
material or by cutting down the factor of safety. 

Another way of solving the problem would have been to assume 
b in terms of P, thus b = cP (c being less than 3), thus giving 

_ faP*( I.26P\ 

In this, substitute trial values for P until one is found which 
satisfies the equation. 

211. Tooth Friction, Pressure and Abrasion.* — From the 
nature of their relative motion there is a sliding, under pressure, 
of the engaging tooth surfaces over each other. The distance 
rubbed over in any case while a pair of teeth are in engagement 
can be computed from the fact that the relative sliding is equal 
to the sum of the addendum of the driver less the engaging length 
of the driven dedendum plus the addendum of the driven tooth, 
less the engaging length of the driver dedendum. This dis- 

* For exhaustive treatment see paper by Lasche in Z. d. V. d. I., Vol. XLIII, 
1899, and by Buchner, Z. d. V. d. I., Vol. XL VI, 1902. Also Bach's Maschinen 
Ekmenle, nth ed., pp. 304-336. 



TOOTHED WHEELS OR GEARS. 379 

tance multiplied by pF' } where f± equals the coefficient of friction 
and F' equals the average normal pressure, gives the friction work 
on each pair of teeth during one engagement. There is no 
difficulty in mathematically determining the space factor for any 
given form and proportions of engaging teeth. A value of p 
depending upon the nature of the surfaces, the lubricant, the 
temperature, whether there is bath lubrication or not, etc., may 
be assumed with a close approximation to the probable actual 
value; but F' will remain a variable term dependent not only 
upon F, the tangential force at the pitch circle contact, but also 
upon the number of teeth simultaneously in action, and the 
form of the path of the point of tooth contact. Experimental 
results show that with accurately cut, properly mounted, well 
lubricated, not overloaded gears this tooth friction loss is so little 
as to be practically negligible. Efficiencies of 99 per cent are 
not unusual. 

The question of the allowable tooth pressure to avoid squeezing 
out of the lubricant and consequent abrasion, or even permanent 
elastic deformation without removal of the lubricant, is one still 
in need of experimental investigation. 

It is evident that the allowable pressure will depend upon the 
properties of the lubricant and the method of its application. 
It is also evident that it will depend upon the radii of curvature 
of the engaging tooth profiles as well as upon the strength, elastic 
yielding and surface hardness of the tooth materials. 

The real factor governing the economic selection of gear- 
tooth materials and proportions will be this one of allowable 
pressure and relative wear rather than the mere ultimate strength. 
Various formulae for allowable pressure have been proposed. 
Lasche * gives charts for rawhide, cast iron, bronze and steel 
gears, which lead to the following equations: 



* Z. d. V. d. I., Vol. XLIII, pp. 1490-1491. 



380 MACHINE DESIGN. 

For cast iron or rawhide, '—^ — : J ^,000. 

exb ^ 

For bronze or steel, ^ — * = 60,000. 

eXb 

F = tangential force at pitch line, pounds; 
r.p.m. = revolutions per minute; 

e = number of pairs of teeth in simultaneous engagement = 
arc of action 
pitch arc 
6 = width of face, inches. 

These equations are, however, only of value for comparable 
conditions — continuous uniform service (motor gears), regularity 
of speed, good workmanship, accurately cut gears, etc. Bach 
proposes the equation for cut, cast-iron gears for continuous 
service 

F=Pb (284.5 -7.iVr.p.m.). 

P = circular pitch, inches. F and b as above. 

This gives values to F which seem too small for well-executed 
installations. It would make F = o, for any cast-iron gears running 
at 1600 r.p.m. or over, which is contrary to experience. The whole 
subject needs further accurate and wide-reaching experimentation. 

212. Non-circular Wheels. — Only circular centrodes or pitch 
curves correspond to a constant velocity ratio; and by making 
the pitch curves of proper form, almost any variation in the velocity 
ratio may be produced. Thus a gear whose pitch curve is an 
ellipse, rotating about one of its foci, may engage with another 
elliptical gear, and if the driver has a constant angular velocity 
the follower will have a continually varying angular velocity. 
If the follower is rigidly attached to the crank of a slider-crank 
chain, the slider will have a quick return motion. This is some- 
times used for shapers and slotting-machines. When more than 



TOOTHED WHEELS OR GEARS. 381 

one fluctuation of velocity per revolution is required, it may be 
obtained by means of "lobed gears"; i.e., gears in which the 
curvature of the pitch curve is several times reversed. If a 
describing circle be rolled on these non-circular pitch curves, the 
tooth outlines will vary in different parts; hence in order to cut 
such gears, many cutters would be required for each gear. Prac- 
tically this would be too expensive; and when such gears are 
used the pattern is accurately made, and the cast gears are used 
without "tooling" the tooth surfaces. 

213. Step, Twisted and Herring-bone Gears. — If a pair of 
wide-faced, ordinary spur gears be divided into several pairs 
of narrow-faced gears and then the successive gears on the one 
shaft be rotated on the shaft by uniformly progressive angles, 
the mating gears will be rotated through corresponding linear 
distances and, therefore, through proportionate angles. Each 
pair of narrow gears will mesh as before, but it is clear that the 
interval of shaft rotation between successive pairs of teeth coming 
into (and going out of) mesh will be reduced in proportion to 
the number of narrow or step gears. This leads to more con- 
tinuous and smooth action. 

If the original gears be considered as made up of a series 
of very thin disks or laminae, the uniform increment of angular 
advance causes each tooth element, originally parallel to the 
shaft axis, to become a helix — the steps having become infinitesimal 
in width. This gives the teeth a spiral form — their profiles in 
a plane normal to their axes of rotation being those, however, 
of the original spur gears. If the twist is in one direction only 
on each gear there will be an end thrust on the shafts when 
the gears are transmitting power, in addition to the regular 
radial and tangential forces. If the gears are made double, or 
herringbone, this end thrust balances itself. Such gears are 
now very accurately cut or hobbed and are applicable to cases 
involving large velocity ratios, high speeds, and fairly large 



382 MACHINE DESIGN. 

amounts of power. It is claimed for them that they are free 
from backlash, vibration and objectionable noise, and very high 
efficiencies (up to 99 per cent) are guaranteed. These gears 
are for connecting parallel axes and must not be confused with 
spiral gears for non-parallel axes. For these see the section 
'" Spiral Gearing." 

214. Bevel-gears. — All transverse sections of spur-gears are 
the same, and their axes intersect at infinity. Spur-gears serve 
to transmit motion between parallel shafts. It is necessary also 
to transmit motion between shafts whose axes intersect. In this 
case the pitch cylinders become pitch cones; the teeth are formed 
b upon these conical surfaces, the result- 
ing gears being called bevel-gears. 
To illustrate, let a and b, Fig. 222, be 
the axes between which the motion is 
to be transmitted with a given velocity 

ratio. This ratio is equal to the ratio 
Fig. 222. ^ 

of the length of the line A to that of B. 

Draw a line CD parallel to a, at a distance from it equal to 

the length of the line A . Also draw the line CE parallel to b, 

at a distance from it equal to the length of the line B. Join 

the point of intersection of these lines to the point O, the 

intersection of the given axes. This locates the line OF, which 

is the line of contact of two pitch cones which will roll together 

MC A 
to transmit the required velocity ratio. For JJr = l>i an0 - lf lt 

be supposed that there are frusta of cones so thin that they may 
be considered cylinders, their radii being equal to MC and NC, 
it follows that they would roll together, if slipping be prevented, 
to transmit the required velocity ratio. But all pairs of radii 

MC J ^ £ 
of these pitch cones have the same ratio, =^, and therefore 

any pair of frusta of the pitch cones may be used to roll together 




TOOTHED IVHEELS OR GEARS. 383 

for the transmission of the required velocity ratio. To insure 
this result, slipping must be prevented, and hence teeth are 
formed upon the selected frusta of the pitch cones. The theo- 
retical determination of these may be explained as follows: 

215. 1st. Cycloidal Teeth. — If a cone, A (Fig. 223), be rolled 
upon another cone, B, their apexes coinciding, an element be of 
the cone A will generate a conical surface, and a spherical sec- 





Fig. 224. 

tion of this surface, adb, is called a spherical epicycloid. Also 
if a cone, A (Fig. 224), roll on the inside of another cone, C, 
their apexes coinciding, an element be of A will generate a conical 
surface, a spherical section of which, bda, is called a spherical 
hypocycloid. If now the three cones, B, C, and A, with apexes 
coinciding, roll together, always tangent to each other on one 
line, as the cylinders were in the case of spur-gears, there will 
be two conical surfaces generated by an element of A : one upon 
the cone B and another upon the cone C. These may be used 
for tooth surfaces to transmit the required constant velocity 
ratio. Because, since the line of contact of the cones is the axo * 
of the relative motion of the cones, it follows that a plane normal 
to the motion of the describing element of the generating cone 
at any time will pass through this axo. And also, since the 
describing element is always the line of contact between the gen- 
erated tooth surfaces, the normal plane to the line of contact, of 
the tooth surfaces always passes through the axo, and the condi- 
tion of rotation with a constant velocity ratio is fulfilled. 

216. 2d. Involute Teeth. — If two pitch cones are in contact 
along an element, a plane may be passed through this element 

* An axo is an instantaneous axis, of which a centro is a projection. 



3^4 



MACHINE DESIGN. 



making an angle (say 75 ) with the plane of the axes of the cones. 
Tangent to this plane there may be two cones whose axes coin- 
cide with the axes of the pitch cones. If a plane is supposed to 
wind off from one base cone upon the other, the line of tangency 
of the plane with one cone will leave the cone and advance in the 
plane toward the other cone, and will generate simultaneously 
upon the pitch cones conical surfaces, and spherical sections of 
these surfaces will be spherical involutes. These surfaces may 
be used for tooth surfaces, and will transmit the required con- 
stant velocity ratio, because the tangent plane is the constant 
normal to the tooth surfaces at their line of contact, and this 
plane passes through the axo of the pitch cones. 

217. Determination of Bevel-gear Teeth. — To determine the 
tooth surfaces with perfect accuracy, it would be necessary to 
draw the required curves on a spherical surface, and then to 
join all points of these curves to the point of intersection of the 
axes of the pitch cones. Practically this would be impossible, 
and an approximate method is used. 

If the frusta of pitch cones be given, B and C, Fig. 225, then 
points in the base circles of the cones, as L, M, and K, will move 
always in the surface of a sphere whose projection is the circle 




Fig. 225. 




LA KM. Properly, the tooth curves should be laid out on the 
surface of this sphere and joined to the center of the sphere to 
generate the tooth surfaces. Draw cones LGM and MHK tan- 



TOOTHED WHEELS OR GEARS, 385 

gent to the sphere on circles represented in projection by lines 
LM and MK. They are called the "back cones." If now 
tooth curves are drawn on these cones, with the base circle 
of the cones as pitch circles, they will approximate the tooth 
curves that should be drawn on the spherical surface. But a 
cone may be cut along one of its elements and rolled out, or de- 
veloped, upon a plane. Let MDH be a part of the cone MHK, 
developed, and let MNG be a part of the cone MGL, developed. 
The circular arcs M D and M N may be used just as pitch circles 
are in the case of spur-gears, and the teeth may be laid out in 
exactly the same way, the curves being either cycloidal or in- 
volute, as required. Then the developed cones may be wrapped 
back and the curves drawn may serve as directrices for the 
tooth surfaces, all of whose elements converge to the center of 
the sphere of motion. 

218. Cutting Bevel-gear Teeth. — The teeth of spur-gears may 
be cut by means of milling-cutters, because all transverse sections 
are alike, but with bevel-gears the conditions are different. The 
tooth surfaces are conical surfaces, and therefore the curvature 
varies constantly from one end of the tooth to the other. Also the 
thickness of the tooth and the width of space vary constantly 
from one end to the other. But the curvature and thickness 
of a milling-cutter cannot vary, and therefore a milling -cutter 
cannot cut an accurate bevel-gear. Small bevel-gears are, 
however, cut with milling-cutters with sufficient accuracy for 
practical purposes. The cutter is made as thick as the narrowest 
part of the space between the teeth, and its curvature is made 
that of the middle of the tooth. Two cuts are made for each 
space. Let Fig. 226 represent a section of the cutter. For the 
first cut it is set relatively to the gear blank, so that the pitch 
point a of the cutter travels -toward the apex of the pitch cone, 
and for the second cut so that the pitch point b travels toward 
the apex of the pitch cone. This method gives an approximation 



386 MACHINE DESIGN. 

to the required form. Gears cut in this manner usually need 
to be filed slightly before they work satisfactorily. Bevel-gears 
with absolutely correct tooth surfaces may be made by planing. 
Suppose a planer in which the tool point travels always in some 
line through the apex of the pitch cone. Then suppose that as it 
is slowly fed down the tooth surface, it is guided along the required 
tooth curve by means of a templet. From what has preceded 
it will be clear that the tooth so formed will be correct. Planers 
embodying these principles have been designed and constructed 
by Mr. Corliss of Providence, and Mr. Gleason of Rochester, 
with the most satisfactory results. 

219. Design of Bevel-gears. — Given energy to be transmitted, 
rate of rotation of one shaft, velocity ratio, and angle between 
axes; to design a pair of bevel-gears. Locate the intersection 
of axes, O, Fig. 227. Draw the axes OA and OB, making the 
required angle with each other. Locate OC, the line of tangency 
of the pitch cones, by the method given on p. 382. Any pair of 
frusta of the pitch cones may be selected upon which to form 
the teeth. Special conditions of the problem usually dictate this 
selection approximately. Suppose that the inner limit of the 
teeth may be conveniently at D. Then make DP, the width 
of face, =DO^-2. Or, if P is located by some limiting condi- 
tion, lay off PD=PO -7-3. In either case the limits of the teeth 
are denned tentatively. Now from the energy and the number 
of revolutions of one shaft (either shaft may be used) the moment 
of torsion may be found. The mean force at the pitch surface 
= this torsional moment -Khe mean radius of the gear; i.e., 
the radius of the point M, Fig. 227, midway between P and D. 
The pitch corresponding to this force may be found. 

In order to compute it consider the teeth of the pinion {i.e., 
the smaller gear), as they will be the weaker. Having found 
the force, F, which is to be transmitted we determine the pitch 
required to carry F by a spur-gear, whose pitch radius =MN 



TOOTHED WHEELS OR GEARS. 



387 



and whose width of face, b, equals PD. (The radius MN is 
used to govern the shape of the tooth and not M R, because the 
teeth are laid off on the developed back cones and not on the 
pitch circles, as has been explained. The circle whose radius 
is MN is sometimes called the formative circle in order to dis- 
tinguish it from the pitch circle.) 

The pitch of such a spur-gear would be the mean pitch of the 
bevel-gearsc But the pitch of bevel-gears is measured at the 




Fig. 227. 

large end, and diametral pitch varies inversely as the distance 
from O. In this case the distances of M and P from O are to 
each other as 5 is to 6. Hence the value of diametral pitch found 
Xf=the diametral pitch of the bevel-gear. If this value does 
not correspond with any of the usual values of diametral pitch, 
the next smaller value may be used. This would result in a 
slightly increased factor of safety. If the diametral pitch thus 
found, multiplied by the diameter 2PQ, does not give an integer 
for the number of teeth, the point P may be moved outward 



388 MACHINE DESIGN. 

along the line OC, until the number of teeth becomes an integer. 
This also would result in slight increase of the factor of safety. 
The point P is thus finally located, the corrected width of face = 
PO -^3, and the pitch is known. The drawing of the gears may be 
completed as follows: Draw AB perpendicular to PO. With A 
and B as centers, draw the arcs PE and PF. Use these as pitch 
arcs, and draw the outlines of two or three teeth upon them, with 
cycloidal or involute curves as required. These will serve to show 
the form of tooth outlines. From P each way along the line AB 
lay off the addendum and the clearance. From the four points 
thus located draw lines toward O terminating in the line DG. 
The tops of teeth and the bottoms of spaces are thus defined. 
Lay off upon AB below the bottoms of the spaces a space HJ 
about equal to \ the thickness of the tooth on the pitch circle. 
This gives a ring of metal to support the teeth. From / draw 
JK toward O. The web L should have about the same thickness 
as the ring has at K. Join this web to a properly proportioned 
hub as shown. The plan and elevation of each gear may now 
be drawn by the ordinary methods of projection. Use large 
fillets, 

220. Twisted Bevel Gears. — In a manner entirely analogous 
to that explained in Sec. 213 concerning the development of 
twisted and herringbone from ordinary spur gears, a pair of 
bevel gears may be considered as made up of very thin en- 
gaging disks or laminae. These may be given progressive 
angles of twist, causing the elements of the teeth, originally 
straight lines converging at the center of relative rotation, to 
become spirals on conical surfaces. Sections on planes nor- 
mal to the axes of rotation remain of the same profile as 
ordinary straight-toothed bevel gears. The same claims of 
improved smoothness of action, noiselessness, high efficiency 
and strength are made for twisted bevel gears as for twisted 
spur gears. 



TOOTHED WHEELS OR GEARS. 



389 



221. Skew Bevel-gears. — When axes which are parallel are 
to be connected by gear-wheels the basic form of the wheels is the 
cylinder. When intersecting axes are to be so connected the basic 
form is the cone or cone frustum. It is sometimes necessary to 
communicate motion between axes that are neither parallel nor 
intersecting. If the parallel axes are turned out of parallelism, 
or if intersecting axes are moved into different planes, so that they 
no longer intersect, the pitch surfaces become hyperboloids of revo- 
lution in contact with each other along a straight line, which is the 
generatrix of the pitch surfaces. These hyperboloids of revolu- 




Fig. 227^, 



tion rotated simultaneously about their respective axes, circum- 
ferential slippage at their line of contact being prevented, will 
transmit motion with a constant velocity ratio. There is, however, 
necessarily a slippage of the elements of the surfaces upon each 
other parallel to themselves. Teeth may be formed on these pitch 
surfaces, and they may be used for the transmission of motion 
between shafts that are not parallel nor in the same plane. Fig. 
227^4 shows, in plan view, a pair of such hyperboloids of revolu- 
tion. Disk portions of these, cut anywhere except at the gorge, 
are approximately conical frusta and are the basic form of skeiv 
bevel-gears. The difficulties of construction and the additional 
■-iction due to slippage along the elements make them undesir- 
ie in practice, and there is seldom a place where they cannot be 




39 o MACHINE DESIGN. 

replaced by some other form of connection. It is evident from 
the figure, for instance, that disk portions taken at the gorge of 
the hyperboloids of revolution are approximately cylinders, which 
are the basic forms of ordinary spiral gears. 

A very complete discussion of skew bevel-gears may be found 
in Reuleaux's "Constructor." 

222. Spiral Gearing.* — If line contact is not essential 
there is much wider range of choice of gears to connect shafts 
which are neither parallel nor intersecting. A and B, Fig. 228, 
are axes of rotation in different planes, both 
planes being parallel to the paper. Let EF and 
GH be cylinders on these axes, tangent to each 
other at the point S. Any line may now be 
drawn, in the plane which is tangent to the 
two cylinders, through S either between A and 
B or coinciding with either of them. This line, say DS, 
may be taken as the common tangent to helical or screw 
lines drawn on the cylinders EF and GH; or helical surfaces 
may be formed on both cylinders, DS being their common tan- 
gent at S. Spiral gears are thus produced. Each one is a por- 
tion of a many threaded screw. The contact in these gears is 
point contact; in practice the point of contact becomes a very 
limited area. 

For the sake of simplicity the special case of spiral gears with 
axes at 90 will be considered first. The term helix angle is here 
taken (as in the treatment of all other screws) as meaning the 
angle of the mean helix with a plane which is perpendicular to 
the axis of rotation. Throughout the discussion the subscript 1 
is used in reference to the driver and the subscript 2 in reference 
to the follower. 

* See further "Worm and Spiral Gearing," by F. A. Halsey. Van Nostrand 
Science Series. Also "Worm Gearing" by H. K. Thomas, McGraw-Hill Book 
Company. 



TOOTHED IVHEELS OR GEARS. 



391 



Let Fig. 228^. represent the plan view of a pair of spiral gears 
with axes at 90 . In this special case it will be noted that the 
axis of the follower lies in the plane perpendicular to the axis of 
the driver and, therefore, that the helix angle of the driver is the 
angle ABC, made by the teeth of the driver with the axis of the 
follower. Call this helix angle of the driver a\. 

Similarly, the helix angle of the follower equals the angle be- 
tween the teeth of the follower and the axis of the driver. Call 
the helix angle of the follower a 2 . 



\ 1 



h- 



1 



;1 



■o^^2f«Z 



I 



r 

Fig. 2 28A 




X 



> 



■^S^ 00^ 



Let Fig. 228$ represent the development of both pitch cyl- 
inders in the tangent plane. It will be noted that the same line 
is normal to the teeth of each gear and that their pitches measured 
on the normal must be equal. The distance occupied by a tooth 
and space on this normal is called the normal pitch. 

If now the driver (1) be moved in the direction AB, the 
follower (2) will be forced to move in the direction CD. For 
a movement of the driver equal to ab, the follower must move a 
distance cb. This establishes the fact that 



circumferential vel. of follower distance moved bv follower 



circumferential vel. of driver distance moved by driver 



cb 
ab' 



T> Cb 

But — = tan a\\ 
ab 



circumferential vel. of follower 
circumferential vel. of driver 



tan «i. (1) 



392 MACHINE DESIGN. 

Let C= distance between centers of gears; 

Di = pitch diameter of driver; 

D 2 = pitch diameter of follower; 

Ni = number of teeth of driver; 

N 2 = number of teeth of follower; 
r.p.m.i = revolutions per minute of driver; 
r.p.m. 2 = revolutions per minute of follower. 

The following equations may then be written: 

circumferential velocity of follower nD 2 r.p.m. 2 , v 

1 . — £ # m f 2 ) 

circumferential velocity of driver nD\ r.p.m. i 
Combining (i) and (2), 

7tD 2 r.p.m.2 

— =tan «i; 

tzD\ r.p.m.i 

D 2 r.p.m.i 
T>\ r.p.m.2 

p-».t t-. / \ ii . r.p.m.2 D\ 

[Note. — Equation (3) may also be written — =-— tan «i. 

r.p.m.i D 2 

r.p.m.2 . . . . 

But is the angular velocity ratio. Hence in spiral gears 

r.p.m.i 

the angular velocity ratio depends upon two factors: first, the 

inverse of the ratio of pitch diameters; second, the tangent of 

the helix angle of the driver. 

In ordinary spur-gears, it will be remembered, the angular 
velocities are inversely as the pitch diameters, or the pitch 
diameters are inversely as the velocities. In spiral gears (axes 
at 90 ) this is only true when tan «i = i, hence when the helix 
angle is 45 . 

Whereas with spur-gears, for a given distance between centers 
and a given velocity ratio, the pitch circles are at once deter- 
mined since there can only be a single pair to fit the conditions; 



TOOTHED WHEELS OR GEARS. 393 

with spiral gears an indefinite number of values may be used 

for D\ and D 2 . It is only necessary to keep =the given 

2 

center distance, and — tan ai = the required velocity ratio. As 
D 2 

D\ and D 2 are varied it is only necessary to vary tan ct\ accord- 

ingly; or, if tan «i is varied — — must be varied accordingly. 

D 2 

For gears with axes at 90 , if «i = 45 9 , a 2 = 45° also, and the 
diameters will be inversely as the angular velocities, i.e., in- 
versely as the revolutions per minute. For all other values of a\ 
this does not hold.] 

Since C= distance between centers 

Di + D 2 =2C; . . . . . . . . (4) 

.\ D 2 =2C-D 1 . ....... (5) 

Substituting in (3) , 

2C—D1 r.p.m.i 

tan «i, 



Di r.p.m.2 

^ ^ ^ /r.p.m.i \ 

2C-Di = Di[ -K : tan a t , 

\r.p.m.2 / 

_ _ / r.p.m.i \ 

2C = DA 1 H tan a\). 

\ r.p.m.2 / 



Di- (6) 

r.p.m.i 

i-\ tan «i 

r.p.m.2 



Equations (6) and (5) will give us all the possible solutions for 

r.p.m.i 
r.p.m.2 



any given values of C and — — — . 



394 



MACHINE DESIGN. 



The particular solution which will best suit a given case is 
determined by practical considerations. 

First, it must be remembered that spiral gears have screw 
action and hence have highest efficiencies for helix angles whose 
value is in the neighborhood of 45 . For satisfactory operation 
it must never be allowed to be less than 15 or over 75 . 

Second, it must be remembered that as a\ approaches 45 
the ratio of the diameters becomes the inverse of the velocity 



Normal Helix 




Fig. 228C. 



ratio. For a very great velocity ratio this may make one of 
the gears too small and the other too large for practical con- 
struction and operation. 

Third, the gears must be such as can be cut with stock 
cutters in any universal milling machines. For this reason 
their normal pitch must have some standard value and the lead 
of the tooth helix must also be a value which can be attained 
with the milling machine. 

The following discussion will deal with this third condition. 
See Fig. 228C, A, which shows a spiral gear extended axially 
to a length sufficient to include a complete tooth-helix. 



TOOTHED IVHFELS OR GEARS. 395 

The normal helix is a helix at right angles to the tooth helix. 
Its entire length is that length which will include (i.e., cut across) 
all the teeth of the gear exactly once. 

Let Li = length of normal helix of driver; 
L2 = length of normal helix of follower ; 
P = normal pitch of each. 
From Fig. 228C, B, which shows a developed pair of gears, 
it can be seen that 

£1 =?!£>! sin ai; (7) 

L2 = 7zD 2 cos ai ; (8) 

.*. —-—-tan a x (9) 

L2 JJ2 

From equation (3), 

r.p.m.2 D\ 

-i- =— tanai, 

r.p.m.x D2 

,_hjj2^ (I0 ) 

L 2 r.p.m.i 

That is, the normal helices are inversely as the number of 
revolutions per minute. Since 

Z,i = length of normal helix of driver, 

P — normal pitch, 
and Ni = number of teeth of driver, 

Li=PNi. 

Similarly, L 2 = PN 2 , 

Li Ni r.p.m.2 . v 

L 2 N 2 r.p.m.i' 

That is, the numbers of teeth are inversely as the numbers of 
revolutions. (Just as in spur-gears.) 





Pp- 


= ?r; 


•'• 


P- 


7Z 






=P; 


•'• : 






■■ p= 






L 2 

N 2 


-P; 


/. 


P = 


N 2 7Z 

L 2 


i 



396 MACHINE DESIGN. 

From (11), N 1 = I ^^N 2 . 

r.p.m.i 

Ni must be a whole number, as must also N 2 \ =— — — iVi ), 

\ r.p.m.2 / 

for a fractional tooth is impossible. Hence a normal pitch 
must be selected which will divide both L\ and L 2 perfectly. 

Also this normal pitch must be a stock value. Let p be 
the diametral pitch corresponding to the normal pitch P. Then 



But 
Also 



and p must correspond to some standard diametral pitch. 

The values of L\ and L 2 as determined by equations (7) 
and (8) may not be such as will give tabular values for p. It 
then becomes necessary to take the nearest standard value for 

P to that obtained from p = —p- and to substitute it in this 

equation, thus deriving a corrected value for L\. Similarly for 
L 2 . This is based upon the assumption that JVi and iV2 are 
given. If p is given, N\ and N 2 must be computed, the nearest 
corresponding whole numbers selected, and L\ and L 2 corrected 
accordingly. 

But changing L\ and L 2 involves also changing D\ } D 2 , 
«i, and a 2 as can be seen by reference to (7) and (8), 

Li = 7iDi sin ai, (7) 

L 2 = nD 2 cos ai, (8) 

(=nD 2 sin <x 2 ). 



TOOTHED WHEELS OR GEARS. 397 

If a\ be not altered, it is evident that D\ and D 2 will be 
altered proportionately with L\ and L 2 and this will change 

the value of the center distance C, since C= . 

2 

If the center distance may be altered to this new value the 
solution is complete. The size of the gear blank of the driver 

2 
would be Di-\ — ; helix angle, «i; number of teeth, Ni ; normal 

P 
(diametral) pitch, p. For the follower the values would be D 2 + 

-; a 2 ( = QO°-ai); N 2 , and p. 
P 

In most cases, however, it will be impossible to change the 
value of C, and the values of Di, D 2 , a±, and a 2 must all be 
changed to keep the corrected values of L\ and L 2 . 

From (7) and (8), 



also, Di-\-D 2 = 2C . . (4) 



Divide by 



£>! = 


Ll D U • 


n sin «i n cos ol\ 




D 1 + D 2 = 2C. . . . 




• Ll 1 u ~C 




71 sin OL\ 7Z COS OL\ 


U 




tz sin ol\ 





L 2 2C . ' 

i+— tan ai = -— Sinai (12) 

Li Li 



Using corrected values of Li and L 2 , try different values for 
a\ until we get an identity. This is the correct value of a\. 
Use this value of a\ and the correct values of Li and L 2 in 

D\ = — 7 and D 2 = ■, 

n sin a\ tz cos a\ 

and solve for corrected values of D\ and D 2 . 



398 



MACHINE DESIGN. 



These corrected values of Di, D 2 , «i r and a 2 (=90° — «i), 
together with N\, N2, and p, already obtained, fully determine 
the gears. 

There remain two points of practical importance to be deter- 
mined: first, the " pitch of the tooth helix;" second, the particular 
cutter of the determined pitch which should be used. 

1. By pitch of the tooth helix is meant the axial length 
corresponding to one complete turn of the tooth helix about 
the pitch cylinder. In ordinary screws this is termed the " lead," 
and as " pitch " is used for so many different purposes we will 
use the term " lead." 

Referring to Fig. 228C, B, it is clear that 

leadi 

— — = tan «i, 

TtD\ 

.'. leadi = 7rDi tan a\. 

lead 2 

= tan a2 = cot a\\ 

71D2 

.'. Iead2 = 7ri^2 cot a\. 

From these leads the gear settings of the milling machine 
are determined. (See Halsey's " Worm and Spiral Gears " 
for table of Brown and Sharpe settings.) 

2. In ordinary spur-gears the cutter to be used for any gear 



*A 



is directly determined by the 
number of teeth of the gear. 
This is not the case with spiral 
gears. 

The method for determining 
the cutter is based upon the 
following reasoning, due to Prof. Le Conte. 

Reference is to Fig. 22&D. a = helix angle as before. 

The figure shows the material of the pitch cylinder extended 



D 


/a 


Ax 


■:!t 


Axis of Gear 








^\ 


V\ / c 






<$>/ 


? 


1 


\ 





Fig. 22&D. 



TOOTHED WHEELS OR GEARS. 399 

either side of the gear; abc is the tooth helix; np represents a 
plane normal to the tooth helix at b. This normal plane will 
cut an ellipse from the pitch cylinder. The minor axis will 
= D, the diameter of the pitch cylinder. The major axis 'is 

D 

determined by the relation — : - = sina; .'. major axis 

major axis 

D 



sin a 

If we cut a spiral gear in the same way as we cut this pitch 
cylinder, selecting the point b midway between two teeth, the 
form of the space on the normal plane will be the true normal 
shape. It will therefore be the true shape of the cutter to be 
used. 

Now the curvature of the normal section of the gear at the 
point indicated is, of course, the curvature of the ellipse at the 
extremity of the minor axis. And the cutter to be used would 
be the cutter for a pitch circle having this curvature. Such a 
circle (i.e., one whose radius equals the radius of curvature of 
the ellipse at the extremity of its minor axis) is called an " oscu- 
lating circle." 

Let p = radius of osculating circle; 

a = half of major axis of ellipse 



2 sin a' 

6 = half of minor axis of ellipse = —. 

2 

D 2 

a 2 4 sin 2 a D 

I hen, p= . = — - — = — — — (12) 

f b D 2 sin 2 a v 0J 



Let iV = number of teeth of normal pitch P on the osculat- 
ing circle, 

••• W.-S (x 4 ) 



400 MACHINE DESIGN. 



Combining (i) and (2), 



No= ^p (i s) 



Let N = actual number of teeth on spiral gear of diameter Z>, 

P' = actual circular pitch of spiral gear. 
Then, 

*-$ O*) 

p 

It will also be seen that — , = sin a, 



P AT 7iD sin a NP , x 

'. P f =~. , .'. N= — , :. «D=— . . 17 

sin a P sin a 



Substituting value of D of (17) in (15), 



NP N 

^0 = -^-^, = — (18) 

sir aP sin* a 



For the driver, then, 



For the follower, 



^0 = ^- (19) 

sm 3 ai 



N =— — = — = — (20) 

sm 6 a 2 cos d ai 



Equations (19) and (20) give the number of teeth whose 
corresponding cutters should be used. 

This completes the solution for spiral gears with axes at 90 . 

The following problem gives the full application of the fore- 
going method. The computation of spiral gears which will run 
together properly calls for strictly accurate numerical work and 
the use of logarithmic tables is recommended. 



TOOTHED W HEELS OR GEARS. 401 

223. Problem. — Design a pair of spiral gears for the follow- 
ing conditions : 

Axes at 90 . £=3.375". 

r.p.m.2 1 revolutions per minute of follower 

r.p.m.i 2 revolutions per minute of driver ' 

D\ = pitch diameter of driver; 

D 2 = pitch diameter of follower; 
a\ = helix angle of driver; 

«2 = helix angle of follower ( = 90°— a 1); 

N\ = number of teeth of driver = 10; 

N 2 = number of teeth of follower = 20; 

Li = normal helix length of driver; 

L 2 = normal helix length of follower. 
It is further assumed that, for reasons of construction, it is 
desired to have the two gears as nearly equal in size as possible. 

First Solution. 
Let D\ = D 2 , and allow C to change in value. 
D 2 r.p.m.i 



D\ r.p.m.2 



tan ai (3) 



2 
i=— tan «i; 



.*. tan «i = J; 
0:1 = 26° 34', .*. 0:2 = 63° 26'. 
Trial L\ = nD\ sin ol\ 

= ^X3.375X.447 2 

= 7TX I.5O9. 

Trial L 2 = 7iD 2 cos 01 

= 7r X3.375X.8944 
= 77X3.018. 



402 



MACHINE DESIGN. 



But p = —p — and N\ has been assumed = 10; 

. . p = = 0.63. 

r 71X1.509 

The nearest standard diametral pitch to this is 7. Selecting 
p = h j ) with Ni = io, 

Actual, or corrected, L\ = N\P= , 

t 

Zi iVi 10 
.. Correct — = — = — =1.429. 

71 p J 

Actual, or corrected, L2 = N 2 P = N 2 -, 

P 



.*. Correct 


u 

71 


_N 2 _ 
' P " 


20 


5 .8 5 8. 




correct L\ 

n sin a\ 






(7) 


= 


1.429 

.4472 



Correct Di = : , (7) = — —=3.195 



2 



Driver gear-blank diameter = Z>H — = 3.i95" + .286" = 3.48i // . 

P 

_ _ correct L 2 ,„ N 2.8^8 

Correct D 2 = , (8) =^- = 3^9S ,f 

71 cos ol\ .8944 

2 

Follower gear-blank diameter = D 2 -\ — = 3. 195" + .286" = 3. 481". 

P 

To select the cutter of the 7 diametral pitch set: 

_ . ,, N\ 10 

Driver Nq = -— — = = 111.8, 

sm d ai .4472 d 

calling for B. & S. involute cutter No. 2. 

Follower N = . , 2 = - = 27.95, 

sm 3 a 2 .8944 s 

calling for B. & S. involute cutter No. 4. 



TOOTHED WHEELS OR GEARS. 



403 



To determine the lead of tooth helix, in order to select the 
corresponding gear set of the milling machine: 

Driver lead = 7rDi tan a\ 

= ttX3.i95X.5 
= 5.018-, 
calling for gears 86, 48, 28, and 100 in B. & S. milling machine. 

Follower lead = 7rZ> 2 cot oi 

= 71X3.195X2 
= 20.075", 

calling for gears 86, 24, 56, and 100 in B. & S. milling machine. 
Summary for modified distance between centers: 



Driver. 

Pitch diameter, ^1 = 3. 195" 
Gear-blank diameter = 3.481" 
Number of teeth, TVi = 10 
Helix angle, 0.1 = 26° 34' 

Diametral pitch, p=j 
Cutter, involute, No. 2 
Lead of tooth helix =5.018" 
Gears, 86, 48, 28, 100 



C 



D! + D< 4 



Follower. 

£2 = 3.195" 

Blank diameter = 3.481" 

iV 2 =20 

02=63° 26' 

Cutter No. 4 

Lead =20.075" 

Gears, 86, 24, 56, 100 

-3-I9S"- 



Second Solution. 

Taking the same data and assuming that the center dis- 
tance remains fixed at 3.375", it is still desired to have D\ 
and D2 as nearly equal as possible. 

If Ni = io, N 2 =2o, and p = y it is fixed that 
£1 = 71X1.429, 
and £2 = 71X2.858. 



404 MACHINE DESIGN. 



T^ / N L>2 2C . 

From equation (12) 1 +— tan 01 = -— sin a h 



6-75 . 

,\ i + 2tanai = sin ai, 

1.429 

or, 1 + 2 tan 01 = 4.724 sin 01. 

The next step is to substitute trial values of a\ until a value 
is found which will make the two sides of the equation equal. 
If the right-hand member comes out greater than the left the 
trial value of 01 is too large; if the left-hand member comes 
out greater than the right the trial value of a 1 is too small. 

Starting with a trial value of 0:1 = 26° 34', from the first 
solution, it is found to be too large. A few trials lead to the 
value of 23° 5' for a\ which gives 

1 + 2 X .42619 = 4.724 X .39207, 
.*. 1.852 = 1.852. 

Therefore the correct 0:1 = 23° 5', and 

correct 02 = 90°— 01 = 66° 55' 

_ „ correct L\ 1.420 

Correct £>i = — ; - = — — = 3.645". 

tc sin 01 (correct) .3921 

n _ correct L 2 2.8 <S 

Correct D 2 = : -= —=3.106". 

n cos 01 (correct) .91993 

2 
Driver gear-blank diameter = B\ + — = 3.645" + .286 = 3.93 1". 

P 

2 
Follower gear-blank diameter = D 2 + — = 3 . 106" + . 286" = 3 -392" 

P 

^ • AT ^1 IO 

Driver iV = -— — = = 165.9, 

sm d oi .392 1 3 

calling for B. & S. involute cutter No. t. 



TOOTHED WHEELS OR GEARS. 405 

N2 20 

Follower N = — — = 5 = 25.1, 

COS d a:i .9199 

calling for B. & S. involute cutter No. 5. 

Driver tooth-helix lead = 7rDi tan a\ 

= ttX3- 645 X. 4262 
= 4-88' v , 
calling for B. & S. gear-set, 48, 64, 56, 86. 

Follower tooth -helix lead = 7rZ> 2 cot a\ 

= ttX 3. 106X2.3463 
= 22.9", 

calling for B. & S. gear-set, 72, 44, 56, 40. 

Summary for fixed distance between centers: 



Driver. 


Follower. 


Pitch diameter, Di = 3.645" 


£2 = 3.106" 


Gear-blank diameter = 3.931" 


Blank diameter = 3.392" 


Number of teeth, N\ = 10 


2V 2 = 2o 


Helix angle 0:1 = 23° 5' 


« 2 = 66° 55' 


Diametral pitch p = y 


P = l 


Cutter, involute No. 1 


Cutter No. 5 


Lead of tooth helix =4.88" 


Lead =22.9" 


Gears, 48, 64, 56, 86 


Gears, 72, 44, 56, 40 


D 1 + D 2 


= 3.2'7C // . 



224. Spiral Gears with Axes at any Angle, |8. — Fig. 228E 
shows a plan view of such a pair of gears, and also a view of the 
gears developed in the tangent plane. 

From the latter it is evident that a motion ba of the driver 
in its direction of rotation must induce a motion be of the 



406 



MACHINE DESIGN. 



follower in its direction of rotation. This establishes the fact 
that 

circumferential velocity of follower be 
circumferential velocity of driver ba' 




Fig. 22&E. 

Consider the triangle abc. Angle cab=a\ ) angle acb — a2, 

be sin a\ 
ba sin a 2 ' 

circumferential velocity of follower 7rD 2 r.p.m. 2 
circumferential velocity of driver izD\ r.p.m.i' 

7rZ) 2 r.p.m.2 sinai 
tzDi r.p.m.i sin^' 

D2 r.p.m.i sin «i 
D\ r.p.m.2 sin a 2 ' 



(1) 



or, 



TOOTHED WHEELS OR GEARS. 407 

r.p.m.2 D\ sin <x\ . . 

— = 7T~' (2) 

r.p.m.i D 2 sin a 2 

Di + D 2 =2C, 

.-. D 2 =2C-D l . ...... (3) 



Substitute in (1) 

2C — D\ r.p.m.i sin a\ 



D\ r.p.m.2 sin a 2 

\r.p.m. 2 sin a 2 ) 

2 c^j 1 (i+ r - p - m - isinai ) 

\ r.p.m.2 sin a 2 )' 



'.'. Di= *—. . ..... (4) 

r.p.m.i sm «i 

r.p.m.2 sin a 2 

Because ai+p + a 2 =i8o°, sin a 2 = sin (J3 + a{) and (4) may 

also be written 

2C 
2>1 = : (5) 

r.p.m.i sin ai 

i-\ — 

r.p.m.2 sin (/? + «i) 

[Note. — Equations (4) or (5) and (3) give us all possible 

solutions for any value of C and -^ — — , just as when the axes 

r.p.m.2 

were at 90 . In fact (5) reduces to the form used in that case 

when #=90°, for 

2C 2C 



Dx = 



r.p.m.i sm a\ r.p.m.i sm a\ 

r.p.m.2 sin (90° + «i) r.p.m.2 cos a\ 

2C 



r.p.m.i 
i-\ tanai. 



r.p.m.2 



4 o8 MACHINE DESIGN. 

It can be shown (Reuleaux's " Constructor ") that the 
sliding velocity (along ac) of the teeth upon each other is least 
when a\ = a2\ .'. whenever possible these values should be 
given ai and a 2 . That is, ai = a 2 = ^(iSo° — /?). 

We have already seen that when /?=9o°, 0:1 = 0:2 = 45° is the 
most efficient angle of helix. 

It must be borne in mind that these values of a'i, a 2 , Di, 
and D 2 may give us impractical values for normal pitch and 
that, in consequence, the values may have to be modified to get 
a normal helix length which will give an even number of teeth 
of stock size.] 

As in spiral gears with axes at 90 we have 

Li = 7iDi sin 01, (6) 

L 2 = nD 2 sin a 2 (7) 

But L 1 = N 1 P=N 1 -. 

P 



L\ N\ lns ( N\ esc a{ 

D\ = — : = — . . 

■k sin a\ p sin «i 



(8) (-^fs) 



Ni = pDismai\ (9) 

n N 2 / N 2 csca 2 \ 

D 2 = ; , ..... (IO) = 

p sin a 2 \ p 1 
N 2 = pD 2 sin a 2 ; (11) 



n n . n Nl csc ^1 + ^2 esc a 2 
2C=Di + D 2 = . . . . (12) 

P 

The foregoing equations may be used to get the practical 
solution. 

225. Problem. — The following problem will illustrate the 
method. Compute a pair of spiral gears, 

Shaft angle, /?=4o°, 



TOOTHED IVHEELS OR GEARS. 

Exact center distance = 3 " (not to be changed), 

r.p.m.i = 400, 

r.p.m. 2 = 3oo, 
p=io. 

Solution. 

(ai + a 2 + $ = i8o , 

ai + a 2 =i8o -4o°=i40 , 



409 



Try 



( 


ai = 


:« 2 = - 1 -=7o °i 

2 


"rom equation (4), 






2C 
D\ — 




2C 6 6 


r.p.m.i sin 

i+— - 

r.p.m.2 sin 


OL\ 

a 2 


— — — — 2A7 , 

r.p.m.i 4 2.333 
1+ 1 + 
r.p.m.2 3 



£> 2 =2C-£>i = 6- 2.57 = 3.43". 
From (9), N\ = pDi sin ai, 

2VTi= 10X2.57 X. 94= 24.1. 
From (ro), N 2 = pD 2 sin a 2 , 

iV2= 10X3.43 X-94 = 3 2 - I 3- 

300 

But N\ and N 2 must be whole numbers in ratio of , or 

400 

24 and 32, respectively. 

Use these values of N\ and N 2 . 
Substitute in (12) 

N\ CSC «i +iViCSCQ:2 

^ = I » 

2/> 



24X1.064 + 32X1.064 

3 = = 2 -979« 

13 2X10 v/y 



4 io MACHINE DESIGN. 

It is evident that new values of a\ and a 2 will have to be tried 
to make this equation an identity. It is necessary to take trial 
values of a\ and a 2 , remembering that the sums of ai and a 2 
must always be i8o°— /? (=140° in this case). 

Try 0:1 = 74°, a 2 = 66°. Substitute in (12) 

24X1 .0403 + 32X1 .0946 
which will answer. 



u /ON , ^ iViCScai 24X1.0403 

From (8), correct D\ = = - = 2.497 

p 10 



// 



iV 2 csca 2 32X1.0946 
From (10), correct Z> 2 = - = = 3-5°3 • 

2 
Driver gear-blank diameter = Z?iH — = 2. 497" + . 2" = 2.697". 

P 

2 
Follower gear-blank diameter = .D 2 4-- = 3.507 // -h2" = 3.703 // . 

P 



Lead of tooth-helix and cutter to be used are found as in spiral 
gears with axes at 90 . 

From the nature of hyperboloidal wheels, two solutions are 
always possible, depending upon whether /? or its supplement 
be taken in determining the line of contact of the hyperboloids. 
In this problem it is evidently just as proper to consider the 
shaft angle to be 140 as 40 . See Fig. 22SF. 

Here 



£+02- 


-OLi)-- 


= 180°, 




a 2 - 


-QLi-- 


= 180°- 


-ft 






= i8o°- 


-140 , 






= 40°. 





TOOTHED WHEELS OR GEARS. 
Try a 2 =7 °> a i = 3°°- 



411 



D x 



2C 



r.p.m.x sin «i 



-3-51^. 



1 + 



r.p.m.9 sin a 2 
J9 2 =2.49 / '. 



SPIRALS 
DRIVER LEFT HAND 
FOLLOWER RIGHT HAND 




Fig. 228F. 



Trial iVi = 17.55, ^2=23.4 (from (9) and (10)), 
Correct, Ni = 1 8, ^2 = 24. 



C= 



N\ esc ai + N 2 esc a\ 



•'■ 3 = 



18X2 + 24X1.064 



2p 20 

showing necessity of modifying a\ and a 2 . 
Try 71 15' and 31 15' for a\ and a 2 , 

18 X 1.9276 + 24 X 1.056 

.*. 3 = = 3.002. 

20 ^ 

Hence these values of a\ and a 2 will answer. 

N\ esc «i 



= 3.0768, 



Correct D\ 



Correct D 2 = 



N 2 esc a 2 

"7" 



3-47", 



2-53 



412 MACHINE DESIGN. 

[It is to be noted that in the two cases the spirals have differ- 
ent relations. In the first case where the spirals are both of 
the same hand, /?+ai + a2=i8o°, i.e., 

ai + a 2 = i8o°-/3. 
In the second case, where the spirals are of opposite hands, 

/?+(a 2 -ai) = i8o , 
i.e., a 2 — ai = i8o° — /?. 

The topic of direction of spirals and direction of rotation 
is well treated in the American Machinist, Oct. n, 1906.] 

226. Worm-gearing. — When the angle between the shafts is 
made equal to 90 , and one gear has only one, two, three, or four 
threads, it becomes a special case of spiral gearing known as 
Worm-gearing. In this special case the gear with a few threads 
is called the worm, while the other gear, which is still a many- 
threaded screw, is called the worm-wheel. If a section of a worm 
and worm-wheel be made on a plane passing through the axis 
of the worm, and normal to the axis of the worm-wheel, the form 
of the teeth will be the same as that of a rack and pinion; in 
fact the worm, if moved parallel to its axis, would transmit rotary 
motion to the worm-wheel. From the consideration of racks 
and pinions it follows that if the involute system is used, the sides 
of the worm-teeth will be straight lines. This simplifies the cutting 
of the worm, because a tool may be used capable of being sharpened 
without special methods. If the addendum equals the reciprocal 
of the diametral pitch, it follows from the interference formula 
that a pressure angle of 75 30' (14J involute system) calls for 
at least 32 teeth on the wheel. For wheels with a smaller number 
of teeth than this, the length of worm addendum must be shortened 
or the pressure angle made smaller {i.e., tooth angle increased). 
If the worm-wheel were only a thin plate the teeth would be 
formed like those of a spur-gear of the same pitch and diameter. 
But since the worm-wheel must have greater thickness, and 
since all other sections parallel to that through the axis of the 



TOOTHED IVHEELS OR GEARS. 



413 



worm, as CD and AB, Fig. 229, show a different form and 
location of tooth, it is necessary to make the teeth of the worm- 
wheel different from those of a spur-gear, if there is to be con- 
tact between the worm and worm-wheel anywhere except in 
the plane EF, Fig. 229. This is accomplished in practice as 
follows : A duplicate of the worm is made of tool steel, and " flutes " 
are cut in it parallel to the axis, thus making it into a cutter, 
which is tempered. It is then mounted in a frame in the same 
relation to the worm-wheel that the worm is to have when they 
are finished and in position for working. The distance between 
centers, however, is somewhat greater, and is capable of being 
gradually reduced. Both are then rotated with the required 
velocity ratio by means of gearing properly arranged, and the 
cutter or "hob " is fed against the worm-wheel till the distance 
between centers becomes the required value. The teeth of the 

CE 





Fig. 230. 

worm-wheel are "roughed out" before they are "hobbed." By 
the above method the worm is made to cut its own worm- 
wheel.* A more modern method is to use a taper hob, fed 
axially. The worm itself is milled, not turned, in many cases. 

The worm may have the basic form corresponding to the 
root circle of the central plane of the wheel. This is known 
in America as the Hindley worm. It gives more teeth in simul- 
taneous engagement than the ordinary cylindrical worm. 

Fig. 230 represents part of the half section of a worm. If it 
is a single worm the thread A, in going once around, comes to 



*This subject is fully treated in Unwin's "Elements of Machine Design, 
and in Brown and Sharpe's "Treatise on Gearing." 



414 MACHINE DESIGN. 

B ; twice around to C, and so on. If it is a double worm the thread 
A, in going once around, comes to C, while there is an inter- 
mediate thread, B. It follows that if the single worm turns 
through one revolution it will push one tooth of the worm-wheel 
with which it engages past the line of centers; while the double 
w r orm will push two teeth of the worm-wheel past the line of 
centers. The single worm, therefore, must make as many revolu- 
tions as there are teeth in the worm-wheel, in order to cause 
one revolution of the worm-wheel; while for the same result 
the double worm only needs to make half as many revolutions 
The ratio of the angular velocity of a single worm to that of 

n 

the worm-wheel with which it engages is=— , in which n equals 

the number of teeth in the worm-wheel. For the double worm 

. . n 
this ratio is — . 
2 

Worm-gearing is particularly well adapted for use where it 
is necessary to get a high velocity ratio in limited space. 

The lead of a worm is measured parallel to the axis of rota- 
tion. The lead of a single worm is P, Fig. 230. It is equal 
to the circular pitch of the worm-wheel. The lead of a double 
worm is Pi = 2P = 2 X circular pitch of the worm-wheel. 

227. Design of Worm-gears. — All spiral gears are forms of 
screw transmission and the formulae for efficiency, etc., developed 
under c, sec. 98, in the chapter on Screws, apply to them di- 
rectly. 

Three points are to be carefully considered in the design of 
worms and wheels : 

1. Speed of rubbing. This is the velocity in feet per minute 
of a point on the pitch line of the worm. The best efficiencies 
are obtained when this is about 200 feet per minute. When 
it exceeds 300 feet there is increasing danger of cutting, and 
the pressure on the teeth must be correspondingly reduced. 



TOOTHED IVHEELS OR GEARS. 415 

At high speeds (say 1000 feet) only very light pressure can be 
sustained without abrasion unless there is bath lubrication. 

2. Pressure on teeth. This depends on the speed and on the 
angle of helix. 

3. Angle of helix. From the formula for screw efficiency we 
have seen that this should be made as great as convenient provided 
it does not exceed 45 . Practical conditions make it impossible 
to use the highest values, but 20 gives very excellent results. It 
should never be less than 15 for fair efficiency. 

Oil-bath lubrication should be used wherever possible; failing 
this, a heavy mixture of graphite and oil has been found satis- 
factory. The following table, based on Professor Stribeck's 
experiments,* applies to a 20 angle of helix and oil-bath lubri- 
cation, using a hardened-steel worm and phosphor-bronze wheel. 

Table XXXIX. 



Rubbing velocity in feet per minute 

Allowable pressure for maximum efficiency 
and continuous operation in pounds 



35oo 



300 
2700 



400 
1850 



500 
1250 



600 



About 60 per cent heavier loads than these were borne, but 
at a loss in efficiency, under continuous operation. For discon- 
tinuous operation, very much heavier loads still are permissible- 
It is largely a question of not allowing the lubricant to reach 
a temperature at which the pressure will rupture the film and 
permit metallic contact. 

This may be taken as a guide. When the angle is greater 
than 20 the values of the pressure may be slightly increased. 
When the angle is less than 20 they should be rapidly diminished; 
thus for io° use only one half the value given. 

There is ordinarily little need to examine the strength of 

* Zeitschrift d. Vereins deutscher Ingenieure, 1897; also 1898. 



4 l6 MACHINE DESIGN. 

worm or wheel teeth. The permissible load (turning force at 
wheel pitch circumference) is limited by questions of number 
of teeth in simultaneous contact, form of tooth profile, nature 
of lubricant and its method of application, allowable rise in 
temperature, etc. It is well to check for the twisting strength 
of the core of the worm. 

Bach and Roser* give the following formula for soft steel 
worm engaging a bronze wheel (helix angle 17 34'; 15 in- 
volute teeth), flooded lubrication. 

W==KPb; 

K=i 4 .2 3 3[ia(h-t)+d]; 

13.17 
a=— — +0.4192; 

21,476 
d = — 24.02. 

F + 541 4V 

W = tangential force at pitch circumference of wheel, pounds ; 
P = circular pitch, inches ; 

b = a,rc length of root of worm-wheel teeth, inches (EF^ 

Fig. 231); 
ti = temperature of oil bath, deg. Fah. ; 

/ = temperature of air, deg. Fah. ; 
V = pitch velocity of worm, feet per minute. 

This is a better formula than Stribeck's value of 

W = 3$6Pb, for cast-iron wheel, 

= 569P6, for phosphor bronze wheel, 
since it is more general. 

Z. d. V. d. L, 1903. 



TOOTHED WHEELS OR GEARS. 4*7 

Cast-iron worms and wheels will run satisfactorily under 
certain conditions,* but a worm of a steel which case-hardens 
well, engaging a wheel of best phosphor bronze, seems to give 
best service. Properly designed and installed worms and wheels 
show high efficiencies, f Particular care must be taken with 
the thrust bearings. 

Since worms and wheels are simply spiral gears in which 
one of the gears has a very few teeth, all of the general formulae, 
relating to D l7 D 2 , «i, ct 2 , p, etc., developed in the preceding 
sections, are directly applicable in their design. However, as 
worms are frequently cut in a lathe (like screw threads) and worm- 
wheels hobbed, it is the axial pitch of the worm, equal to the 
circumferential pitch of the wheel, which is of importance rather 
than the normal pitch. This " lead " must then have a value 
obtainable with the screw-cutting gearing of the lathe. It is 
therefore practically more convenient in the design of worms 
and wheels to follow the method illustrated by the following 
problem : 

228. Problem. — Two shafts about 10 inches apart and at a 
right angle with one another are to have a velocity ratio of 20 to 1, 
The worm-shaft makes 300 revolutions per minute. 

Since the velocity ratio is 20 to 1, the wheel will have to have 
20, 40, or 60 teeth, depending upon whether the worm is single, - 
double-, or triple-threaded. 

If the shafts are 10 inches apart the greatest allowable pitch 
radius of the wheel will not be far from 8 inches; 50 inches may 
be taken as a trial pitch circumference of the wheel. 

With a single-thread worm this will give a circular pitch of 
-f$ = 2§ inches. With a double thread the circular pitch would 
be f$=i \ inches. 

* Stribeck. Z. d. V. d. I., 1898. 

f See further Kenerson, Trans. A. S. M. E., Vol. XXXIV. Also Bruce, Proc. 
Inst. M. E., 1905. 



418 MACHINE DESIGN. 

In any case the rise of the pitch helix of the worm will be 
2 J inches for one revolution. 

This value must always be such that the thread may be cut in 
an ordinary lathe. 

If it is required that the helix angle, a, be 20 , then the 
pitch circumference of the worm must be such that 

tan 2q0 = ±5 

°* pitch circumference of worm ; 

2.5 
.*. pitch circumference of worm = — — =6.87 inches. 

0.364 

Pitch diameter of worm=— : — =2.2 inches. 



50 
Pitch diameter of wheel = — = 1^.88 inches. 

15.88 + 2.2 

Actual distance between shafts = == 9-o4 inches. 



The question now arises whether 2.2 inches is a great enough 
pitch diameter for the worm. If the thread is single the pitch 
= 2.5 inches and the corresponding dedendum=o.92 inch. 

Twice this dedendum='i.84 inches, which subtracted from 
2.2 inches would only leave a central solid core of 0.36 inch diam- 
eter for the worm. It is obvious without computation that this 
would not sustain the torsional moment. If the double thread 
were used the central core would have a diameter of 1.28 inches. 

For each revolution of the worm the length of the path of 
the point of contact or the distance rubbed over equals the helix 
length on the pitch line. This is the hypothenuse of a right- 
angle triangle whose base =6.87 inches and whose altitude = 
2.5 inches, or 7.3 inches. 

At 300 revolutions per minute the distance rubbed through in 



TOOTHED WHEELS OR GEARS.] 4*9 

feet per minute =300 X — = 182 feet. At 182 feet the allowable 

pressure, W, between the teeth may equal 3500 lbs., assuming 
bath lubrication, a steel worm, and a bronze wheel This is the 
pressure applied at the circumference of the woim-wheel in the 
direction of the axis of the worm. The total work done on the 
worm-wheel in foot-pounds per minute will equal W multiplied 
by the pitch velocity of the wheel in feet per minute. 

This wheel makes Vo 9 " = 1 S revolutions per minute and its 
pitch circumference =f| feet, hence its pitch velocity = 15 X^|- 
= 62.5 feet per minute. 

3500X62.5 =218,750 ft. -lbs. =6.63 H.P. 

This same amount of energy is transmitted through the worm. 
The twisting moment on the shaft = Fr, where r equals the pitch 
radius of the worm. F = energy transmitted -f- the velocity of 
the point of application of the force. 

^ 218750 „ 

-1274 lbs.; 



>- 8 7 
300 v 



12 

i*V= 1274X1.1 = 1401 in. -lbs. 
To resist this there is a circular section whose strength is 
represented by/ s 



nri 3 



2 

ri = radius of core of worm = 0.64 inch; 

/s = unit stress in outer fiber; 

Fr 1401 

. . /«= — »= = 34oo lbs. 

ixy\ 1 .412 

2 

This is a safe value for steel. Therefore the double-threaded 
worm will be used and the wheel will have 40 teeth of i\ inches 
circular pitch. 



420 MACHINE DESIGN. 

Had the distance between the shafts been fixed at 10 inches 
the helix angle could not have been assumed but must have been 
calculated. 

The pitch radius of worm would have been 

IO „_ii§£ =2 . 02 „ 



Pitch circumference = 12.69 inches. 
Tangent of helix angle = — ^-=0.1955. 

.". a = n° + . 

With the center distance fixed at 10 inches, the helix angle 
need not necessarily be as low as n°; provided, of course, that 
the axial pitch of the worm may be changed to some other value 
greater than 2. 5". As a check on the final results the general 
formulae of the preceding sections may be applied to the values 
obtained by the method here followed. 

When the worm and worm-wheel are determined, a working 
drawing may be made as follows: Draw AB, Fig. 231, the axis 
of the worm-wheel, and locate O, the projection of the axis of 
the worm, and P, the pitch-point. With O as the center draw the 
pitch, full depth, and addendum circles, G, H, and K; also 
the arcs CD and EF, bounding the tops of the teeth and the 
bottoms of the spaces of the worm-wheel. Make the angle ($ = 90°. 
Below EF lay off a proper thickness of metal to support the teeth 
and join this by the web LM to the hub N. The tooth outlines 
in the other sectional view are drawn exactly as for an involute 
rack and pinion. Full views might be drawn, but they involve 
difficulties of construction, and do not give any additional infor- 
mation to the workman. The drawing should contain a clear 
statement of the size and form of the worm tooth, the lead, 
whether the worm is single, double, triple, or quadruple threaded, 



TOOTHED WHEELS OR GEARS. 



421 



the number of teeth of the wheel, and its helix angle, in addition 
to all ordinary dimensions. 




Fig. 231. 

229. Compound Spur-gear Chains. — Spur-gear chains may 
be compound, i.e., they may contain links which carry more than 
two elements. Thus in Fig. 232 the links a and d each carry 
three elements. In the latter N case the teeth of d must be counted 
as two elements, because by means of them d is paired with both 
b and c. In the case of the three-link spur-gear chain, Fig. 197, 
the wheels b and c meshed with each other, and a point in the 
pitch circle of c moved with the same linear velocity as a point in the 
pitch circle of b, but in the opposite sense. In Fig. 232 points in 
all the pitch circles have the same linear velocity, since the motion 
is equivalent to rolling together of the pitch circles without slip- 
ping; but c and b now rotate in the same direction. Hence it 
is seen that the introduction of the wheel d has reversed the 
direction of rotation, without changing the velocity ratio. The 



422 



MACHINE DESIGN. 



size of the wheel, d, which is called an " idler," has no effect upon 
the motion of c and b. It simply receives upon its pitch circle 
a certain linear velocity from c, and transmits it unchanged 
to b. Hence the insertion of any number of idlers does not 
affect the velocity ratio of c to b, but each added idler reverses 
the direction of the motion. Thus, with an odd number of 
idlers, c and b will rotate in the same direction; and with an 
even number of idlers c and b will rotate in opposite directions. 

If parallel lines be drawn through the centers of rotation of 
a pair of gears, and if distances be laid off from the centers on 
these lines inversely proportional to the angular velocities of the 
gears, then a line joining the points so determined will cut the 
line of centers in a point which is the centro of the gears. In Fig. 
232, since the rotation is in the same direction, the lines have to 




Fig. 232. 



Fig. 233. 



be laid off on the same side of the line of centers. The pitch 
radii are inversely proportional to the angular velocities of the 
gears, and hence it is only necessary to draw a tangent to the pitch 
circles of b and c, and the intersection of this line with the line of 
centers is the centro, be, of c and b. The centrodes of c and b are 
C\ and bit circles through the point be, about the centers of c and b. 
Obviously this four-link mechanism may be replaced by a three- 
link mechanism, in which ci is an annular wheel meshing with a 
pinion b\. The four link mechanism is more compact, however, 
and usually more convenient in practice. 

The other principal form of spur-gear chain is shown in 



TOOTHED WHEELS OR GEARS. 423 

Fig. 233. The wheel d has two sets of teeth of different pitch 
diameter, one pairing with c and the other with b. The point 
bd now has a different linear velocity from cd, greater or less in 
proportion to the ratio of the radii of those points. The angular 
velocity ratio may be obtained as follows: 

angular velocity d _ C . . . cd 
angular velocity c D . . . cd* 



also, 

Multiplying, 



angular velocity b __ D . . .bd 
angular velocity d B . . . bd' 

angular velocity b C . . . cdxD . . . bd 



angular velocity c D . . . cdxB . . . bd' 



The numerator of the last term consists of the product of the 
radii of the "followers," and the denominator consists of the 
product of the radii of the "drivers." The diameters or numbers 
of teeth could be substituted for the radii. 

In general, the angular velocity of the first driver is to the 
angular velocity of the last follower as the product of the number 
of teeth of the followers is to the product of the number of teeth 
of the drivers. This applies equally well to compound spur-gear 
trains that have more than three axes.* Therefore, in any spur- 
gear chain the velocity ratio equals the product of the number of 
teeth in the followers divided by the product of the number of 
teeth in the drivers. The direction of rotation is reversed if the 
number of intermediate axes is even, and is not reversed if the 
number is odd. If the train includes annular gears their axes 
would be omitted from the number, because annular gears do 
not reverse the direction of rotation. 

A common modification of the chain of Fig. 233 is shown in 
Fig. 233 A. Here the axis of the gear c is made to coincide with 

* Epicyclic trains excepted. 



424 



MACHINE DESIGN. 



the axis of b, and the mechanism is known as a reverted gear train. 
Probably the best known application of this mechanism is that of 
the backgearmg of the ordinary engine lathe. The velocity ratio 
of c and b is, of course, not altered by having their axes coincide, 
and it is equally evident that one of them only may be keyed to 
the shaft while the other is free to rotate on it. 





Fig. 233 A. 



Fig. 233 B. 



230. Epicyclic Gearing. — In the gear trains of the preceding 

sections, the velocity ratios have been studied with reference to 

a fixed member to which each gear is attached by a turning pair. 

Fig. 233 B illustrates such a simple chain of three links, a, b, and 

c. Considering a as fixed it is evident that, if c makes m turns 

per minute relatively to a, causing b to make n turns per minute 

ft 
relatively to a, for one turn of c relatively to a, b will make — turns 

m 

relatively to a. The ratio — is called the velocitv ratio, and is 

m " 

designated by r. 

If, now, one of the gears, c, be considered as the fixed link, 
and it is desired to examine the action of the mechanism when 
a is swung about ac as center, it is evident that a different 
mechanism is obtained. See §8 and § 12. The action can be 
explained under the general laws laid down in these sections 
but can be understood more readily by reference to Fig. 233 C- 
Such mechanisms are known as epicyclic gear trains, because 
points in the one gear describe epicycloidal curves relatively to 
the other gear. The name has no connection with the form of 
the gear teeth which may belong to the cycloidal, involute, or 
any other system. 



TOOTHED WHEELS OR GEARS. 



425 



Let it be supposed that the three links can be rigidly locked 
together and while so held are given a complete turn about the 
axis ac, in a clockwise direction. Owing to this, b will make 
one turn in a clockwise direction about its own axis ab. In 
position 1 the arrow is seen to be horizontal, and to the left of ab, 
at 2 it is vertical and above ab, at 3 horizontal and to the right, 
at 4 vertical and below, and at 1, when the turn about ac has 
been completed, it is once more horizontal and to the left of ab< 




£>— >)3 



Fig. 233 C. 

The arrow on b has, therefore, made a complete turn about ab 
as axis, and if one line of the rigid body b has made such a turn 
the whole body b has done so. But in swinging the locked 
mechanism about ac, the link c has been given a complete revo- 
lution in a clockwise direction. This is contrary to the original 
assumption that c be the fixed link, i.e., remain at rest. If, 
now, the mechanism be unlocked and c be given a complete 
revolution in a counter-clockwise direction while a is held sta- 
tionary, the result will be the same as though c had not been 
allowed to move at all. But this counter-clockwise revolution 
of c will cause b to- have a further clockwise rotation about its 

axis of — - r turns. The total number of turns which b makes 
m 

about its axis while a makes one turn about ac will, therefore, 
equal 1 + r 



426 



MACHINE DESIGN. 



Had an idler been placed between b and c, the result would 
have been to cause b to be given r turns in a counter-clockwise 
or negative direction, when c was brought back to its original 
position and, consequently, b would make i—r revolutions about 
its axis for one revolution of a about ac. This can be seen 
clearly in Fig. 233 D, where b and c are purposely made the 




Fig. 233 P. 

same size so that r =1 and, hence, 1 — r =0. In other words, 
in this special case the gear b does not rotate about its axis 
at all; its motion, as can be seen from positions 1, 2, 3, and 4, 
being merely translation, as the arrow on b remains always parallel 
to its original position. 

A second intermediate gear, or idler, would again reverse 
the direction of 5's motion, making the revolutions of b =i+r. 

The general law may be stated as follows: — "The number of 
revolutions made by the last wheel of an epicyclic train for each 
revolution of the arm is equal to the one plus the velocity ratio 



TOOTHED WHEELS OR GEARS. 4 2 7 

of the train if the number of axes in the train be even, and one 
minus the velocity ratio of the train if the number of the axes be 
odd. In the former case the wheel turns in the same sense as 
the arm ; in the latter in the opposite sense, unless the ratio r 
is less than unity." (Kennedy — Mechanics of Machinery.) 

The same holds if there are no annular gears in the train or 
if there are an even number of them. If, however, there be one 
or any other odd number of annular gears in the train, the effect 
will be to transpose the plus and minus as well as the sense of 
rotation. 

If the first wheel of any epicyclic train has its axis fixed, but 
has itself a motion of rotation about this axis so that, for example, 
it makes k revolutions for each revolution of the arm, then the 
last wheel of the train will make i±r±kr revolutions instead 
of i ±r. The sign of r is determined as before but the sign of 
kr is plus, if the rotation of the first wheel causes the last wheel 
to rotate in the same sense as the arm, and minus, if the rotation 
of the first wheel causes the last w^heel to rotate in a sense opposite 
that of the arm. 

The only case which requires special attention for fear of 




Fig. 233 E. 

incorrectly determining the number of axes is where the gear 
train of Fig. 233, which has three axes, is given the reverted form 
shown in Fig. 233 E, which apparently has but two axes. For 
proper analysis it is necessary to consider the reverted train 
the same as the original form, i.e., a double axis is counted as 
two single ones in computing the number of axes in the train. 



428 MACHINE DESIGN. 

Problem. — Find the number of revolutions c will make about 
its axis for each revolution of the arm a; d being considered as 
the fixed link. 

d has 101 teeth and meshes with b which has ioo teeth, b' is 
keyed to same shaft as b, has 99 teeth, and meshes with c> which 
has 100 teeth. . If this were an ordinary reverted gear train with 
a as fixed link, then, remembering that the angular velocity of 
the last follower is to the angular velocity of the first driver as 
the product of the number of teeth of the drivers is to the product 
of the number of teeth of the followers, for one turn of d, c would 

make -^- = WW turns in the same sense. This is r, the 

100 X 100 1 0000 

velocity ratio of the train. Considering the train as an epicyc- 
lic one with d as fixed link, there are three axes and no annular 
gears and the rule would be that for one turn of a in a clock- 
wise direction c would make 1 — r turns about its axis in 

1 0000 1 
the same sense, equal to 1 — = • 

TCCOO IOOOO 



CHAPTER XVIII. 

SPRINGS. 

231. Springs Defined. — Usually machine members are required 
to sustain the applied forces without appreciable yielding and are 
designed accordingly; but certain machine members are useful 
because of considerable yielding. They are generally called springs. 

232. Illustrations. — (a) The spring of a safety-valve on a steam- 
boiler holds the valve down until the steam -pressure reaches the 
maximum allowable value; then it yields and allows steam to 
escape until the pressure is reduced, when it closes the valve. 

(b) The springs upon which a locomotive-engine is supported 
prevent the transmission of the full effect of the shocks, due to 
running, to the working parts of the engine, thereby reducing the 
resulting stresses. Car-springs in a similar manner protect 
passengers and freight. 

(c) "Bumper" springs reduce stresses in cars and their con- 
tents due to axial shocks. 

(d) The springs in certain steam-engine governors yield under 
the increased centrifugal force of the governor weights, due to 
increased rotative speeds, and allow the adjustment of the valve- 
gear to the changes of effort and load. 

(e) Heavy reciprocating parts are often brought to rest without 
shock and are then helped to start on their return travel by the 
expanding spring. 

(/) A power-hammer strikes a "cushioned blow" because of the 
action of a spring. This spring may be of steel, rubber, or steam. 

(g) Belt connections are really yielding members and tend 
to reduce shocks transmitted through them; while gears (except 
rawhide or " hard- fiber ") yield almost imperceptibly and trans- 
mit shocks almost unchanged. 

429 



43° 



MACHINE DESIGN. 



(h) Long bolts may become springs for the reduction of 
stress due to shock. 

(i) Springs may serve for the storing of energy which is 
given out slowly to actuate light-running mechanisms, like clocks. 

233. Cantilever Springs. — Many springs are simple cantilevers 
with end loads. (See Fig. 234.) 



^ 1 . 



Fig. 234. 



Fig. 235. 



The rectangular spring of constant width b and height (or 
thickness) h, with a load F applied at a distance / from the sup- 
port gives, from the laws of beams, 



F = 



fM 2 



and A = 



4FP 



61 " Ebh*' 

f b is the unit stress in the outer fiber in pounds per square inch, 

all forces being expressed in pounds and all dimensions in inches ; 

A is the total deflection in inches due to the application of F; 

E is the modulus of elasticity of the material used. 

FA f b 2 V 
The work done or energy stored = — = — — , where V= volume, 

2 10E 

bhl, in cubic inches. 

For a flat spring of uniform breadth b, rectangular cross- 
section, top surface flat and lower surface a parabola in outline, 
such as is shown in Fig. 235, 



F= 



61 



and A = 



6FP 

EbW 



Work done or energy stored =' 



12'E' 



SPRINGS. 



43 1 



The same equations hold approximately for the cantilever 
spring shown in Fig. 236. They also hold for the triangular 
spring of constant depth h shown in Fig. 237. 









I 


* -t » 


y a 


> 


F 






Fig. 236. 



Fig. 237. 



In all of these cases obviously the yielding varies inversely 
as h 3 , and the strength directly as h 2 \ hence, if h be increased 
to obtain required strength, the yielding will be decreased as the 
cube of h while the strength is increased only as the square of h. 
Much of the requisite yielding is therefore sacrificed if the strength 
is obtained by increasing h. 

Inspection of the same equations shows that increasing the 
breadth b to obtain the required strength decreases the deflection 
in the same proportion. In springs, 
therefore, where yielding is to be kept 
large, it is better to gain requisite 
strength by varying b; while in a beam 
h should be as large as possible because 
here deflection is to be reduced to the 
smallest value. If the spring is to be of 
tool steel, hardened and tempered, thin 
material is better suited to the operation 
of hardening. 

As b is increased with a constant 
small value of h, it may become too great for the available 
space. This difficulty is overcome as shown by reference to 
Fig. 238. 

Suppose ABC is a triangular spring designed for certain con- 




Fig. 238. 



43 2 



MACHINE DESIGN. 



ditions of load and yielding. AB is an inconvenient width. 
Divide AB into equal parts, say six. Conceive the portion 
GFHBi cut off and placed in the position B x LKiE u and simi- 
larly conceive that EDKAi occupies the position AiMKiEi, 
and the two parts are rigidly joined along the line K i E 1 . Also 
conceive the portion ADE moved to A1D1E1, and BFG moved 
to B1D1E1, and that they are rigidly joined along the line D\Ei. 

The amount of material is unchanged. The bending force is 
applied in nearly the same way to the portions whose position is 
changed. The leaf spring is therefore practically equivalent 
to the triangular spring from which it is made. 

The following equations are given by Prof. J. B. Peddle,* 
for leaf springs having both full and pointed leaves of equal base 
number of full length leaves 



width. Let r 



total number of leaves 




r — * — j I r — l — H 



Fig. 2384. Fig. 2 3 85. 

For semi-elliptic spring, Fig. 238A : 

fbW 

"*' 

b = number of leaves X width per leaf, inches; 

h= thickness per leaf, inches; 

/=unit stress in outer fiber, pounds per square inch. 

* American Machinist, April 17, 1913. Prepared for Halsey's "Handbook for 
Machine Designers." 



SPRINGS. 

hE ' 
For full elliptic spring, Fig. 2 3 SB. 

fW 

4 p/k 

K= 



433 



(1 



1 -r 2 I 
2r (1 -r) -r 2 \og e r . 



234. Springs for Axial Loads. — Many springs are subjected 
to axial loads; they are usually helical in form as shown in 
Fig. 239. 





(t- 



Fig. 239. 



F may act to stretch or compress the spring. 
Consider the cross- section of the rod to be circular. 
Let F=load in pounds; 

d= diameter of rod in inches; 
a = mean radius of coil in inches; 
iV=number of coils; 
/= developed length of spring in inches = 2naN\ 
/,= allowable unit shearing stress in outer fiber in pounds 

per square inch ; 
£,=modulus of shearing elasticity =|E; 
A = extension or compression in inches. 
Then the following equations may be developed : 



zji6Fa 



22Fa 2 l Ad±E s f 2 V 

4-2-jZT, and 2V=— ± Work done =^-. 

fa nd*E s 64F& 4b, 



434 



MACHINE DESIGN. 



In a helical spring for an axial load using a rectangular cross- 
section of wire the axial height of wire = h and radial breadth = b y 
the equations become 



p = lt 



b 2 h 2 



$a ^tf + h 2 ' 



J Fa 2 l b 2 + h 2 , „ AE R b 3 h? 

A = 3 -pr- • r.,,0 , and N = 



£, b*h 3 



6nFa? b 2 + h 2 ' 



It will take one and a half times as much material to make a spring 
of this type as it would to make a round-wire helical spring of 
equal strength. 

235. Springs for Torsional Movements. — Many springs come 




Fig. 240. 



under class (i) as mentioned above. The general case is shown 
in Fig. 240. The spring is a spiral of flat wire with an axial 
height b and radial depth h. 

F= turning force in pounds; 

a = lever-arm in inches ; 

/ 6 = unit stress in outer fiber, pounds per square inch; 

/= developed length of spring; 

$ = angular deflection ; 

J = distance inches moved through by the point of appli- 
cation of F. 



Then 



fM 2 
6a 



and d = a& 



\2Fla 2 
Ebh? ' 



fb 2 V 
Work done or energy stored = — — . 

6E 

236. Materials and Allowable Stresses. — Springs may, of 
course, be made of any material having elastic strength; but 



SPRINGS. 435 

spring steel is the material most frequently used. In its un- 
tempered or soft condition it has an elastic limit of from 70,000 
to 90,000 lbs. per square inch. Tempered, or hard drawn, its 
elastic limit may be from 110,000 to 140,000 lbs. per square 
inch. 

A unit bending stress of 80,000 lbs. and a shearing stress of 
60,000 may be used with safety. 

£ = 30,000,000 and £5 = 12,500,000. 

For round rod torsional springs the following values may 
be used. 



Diam. of wire, in. 


Safe unit stress, f s , 


1 

4. 


70,000 


1 


60,000 


i 


50,000 



For further information the reader is referred to Reuleaux's 
« Constructor," Trans. A. S. M. E., Vols. V and XVI, and Hal- 
sey's u Handbook for Machine Designers." 



CHAPTER XIX. 

MACHINE SUPPORTS. 

237. General Laws for Machine Supports. — The single-box 
pillar support is best and simplest for machines whose size and 
form admit of its use. When a support is a single continuous 
member, its design should be governed by the following principles : 

I. The amount of material in the cross -section is determined 
by the intensity of the load. If vibrations are also to be sus- 
tained, the amount of material must be increased for this purpose. 

II. The vertical center line of the support should coincide 
with the vertical line through the center of gravity of the part 
supported. 

III. The vertical outlines of the support should taper slightly 
and uniformly on all sides. If they were parallel they would 
appear nearer together at the bottom. 

IV. The external dimensions of the support must be such 
that the machine has the appearance of being in stable equilibrium. 
The outline of all heavy members of the machine supported must 
be either carried without break to the foundation, or if they 
overhang, must be joined to the support by means of parabolic 
outlines, or by the straight lines of the brace form. 

238. Illustration. — In Fig. 241 the first three principles may 

be fulfilled, but there is an appearance of instability. It is 

because the outline of the "housing" overhangs. It should be 

carried to the foundation without break in the continuity of the 

metal, as in Fig. 242. 

436 



MACHINE SUPPORTS. 



437 



239. Divided Supports. — When the support is divided up into 
several parts, modification of these principles becomes necessary, 
as the divisions require separate treatment. This question may 




Fig. 24 t. Fig. 242. 

be illustrated by lathe supports. In Fig. 243 are shown three 
forms of support for a lathe, seen from the end. For stability 
ihe base needs to be broader than the bed. In A the width of 
base necessary is determined and the outlines are straight lines. 
The unnecessary material is cut away on the inside, leaving legs 
which are compression members of correct form. The cross - 
brace is left to check any tendency to buckle. For convenience 
to the workmen it is desirable to narrow this support somewhat 




Fig 243. 

without narrowing the base. The cross-brace converts the single 
compression member into two compression members. It is allow- 
able to give these different angles with the vertical. This is done 
in B and the straight lines are blended into each other by a curve. 
C shows a common incorrect form of lathe support, the compres- 
sion members from the cross-brace downward being curved. 
There is no reason for this curved form. It is less capable of 



438 



MACHINE DESIGN. 



bearing its compressive load than if it were straight, and is no 
more stable than the form B, the width of base being the same. 




V 



'y/2& 



Fig. 244. 



Consider the lathe supports from the front. Four forms are 
shown in Fig. 244. If there were any force tending to move the 
bed of the lathe endwise the forms B and C would be allowable. 
But there is no force of this kind, and the correct form is the 
one shown in D. Carrying the foot out as in A, B, and C in- 
creases the distance between supports (the bed being a beam 
with end supports and the load between); this increases the de- 
flection and the fiber stress due to the load. This increase in 
stress is probably not of any serious importance, but the prin- 
ciple should be regarded or the appearance of the machine will 
not be right. If the supports were joined by a cross-member, 





Fig '45. 

as in Fig. 245, they would be virtually converted into a single 
support, and should then taper from all sides. 

240. Three-point Support. — If a machine be supported on 
a single -box pillar, change in the form of the foundation cannot 
induce stress in the machine frame tending to change its form. 
If, however, the machine is supported on four or more legs the 



MACHINE SUPPORTS. 439 

foundation might sink away from one or more of them and leave 
a part unsupported. This might cause torsional or flexure stress 
in some part of the machine, which might change its form and 
interfere with the accuracy of its action. 

But if the machine is supported on three points this cannot 
occur, because if the foundation should sink under any one of 
the supports the support would follow and the machine would 
still rest on three points. When it is possible, therefore, a ma- 
chine which cannnot be carried on a single pillar should be sup- 
ported on three points. Many machines are too large for three- 
point support, and the resource is to make the bed, or part sup- 
ported, of box section and so rigid that even if some of the legs 
should be left without foundation the part supported would still 
maintain its form. More supports are often used than are necessary. 
Thus, if a lathe has two pairs of legs like those shown in B, 
Fig. 243, and these are bolted firmly to the bed, there will be 
four points of support. But if, as suggested by Professor Sweet, 
one of these pairs be connected to the bed by a pin so that the 
support and the bed are free to move relatively to each other 
about the pin, as in Fig. 246, then this is equivalent to a single 
support, and the bed will have three points of support, and will 
maintain its form independently of any change in the foundation. 
This is of special importance when the machines are to be placed 
upon yielding floors. 

241. Reducing Number of Supports. — Fig. 247 shows another 




«/---» A --vf-":-Y 



Fig. 247. 

case in which the number of supports may be reduced without 
sacrifice. In A three pairs of legs are used. There are therefore 



440 



MACHINE DESIGN. 



six points of support. In B two pairs of legs are used and one 
may be connected by a pin, and there will be but three points of 
support. The chance of the bed being strained from changing 
foundation has been reduced from 6 in A to o in B. The total 
length of bed is 12 feet, and the unsupported length is 6 feet 
in both cases. 

242. Further Correct Methods. — Figs. 248 and 249 show 
correct methods of support for small lathes and planers, due to 



T^i 



Fig. 248. 




Fig. 249. 



Professor Sweet. In Fig. 248 the lathe "head-stock " has its 
outlines carried to the foundation by the box pillar; a represents 
a pair of legs connected to the bed by a pin connection, and 
instead of being placed at the end of the bed it is moved in some- 
what, the end of the bed being carried down to the support by a 
parabolic outline. The unsupported length of bed is thereby 
decreased, the stress on the bed is less, and the bed will maintain 
its form regardless of any yielding of the floor or foundation. 
In Fig. 249 the housings, instead of resting on the bed as is usual 
in small planers, are carried to the foundation, forming two of 
the supports; the other is at a and has a pin connection with the 
bed, which being thus supported on three points cannot be twisted 
or flexed by a yielding foundation. 



CHAPTER XX. 



MACHINE FRAMES. 



243. Open-side Frame. — Fig. 250 shows an open -side frame, 
such as is used for punching and shearing machines. During the 
action of the punch or shear a force is applied to the frame tending 
to separate the jaws. This force may be represented in magnitude, 
direction, and line of action by P. It is required to find the 
resulting stresses in the three sections AB, CD, and EF. Con- 
sider AB. Let the portion above this section be taken as a free 
body. The force P, Fig. 251, and the opposing resistances to 




Fig. 250. 

deformation of the material at the section AB, are in equilibrium. 
Let H be the projection of the gravity axis of the section AB, 
perpendicular to the paper. Two equal and opposite forces, Pi 
and P 2 , may be applied at H without disturbing the equilibrium. 
Let Pi and P2 be each equal to P, and let their line of action be 
parallel to that of P. The free body is now subjected to the 
action of an external couple, PI, and an external force, P\. The 
couple produces flexure about H, and the force P\ produces tensile 
stress in the section AB. The flexure results in a tensile stress 
varying from a maximum value in the outer fiber at A to zero 

441 



442 



MACHINE DESIGN. 



at H, and a compressive stress varying from a maximum in the 
outer fiber at B to zero at H. This may be shown graphically 
at JK. The ordinates of the line LM represent the varying 
stress due to flexure; while ordinates between LM and NO 
represent the uniform tensile stress. This latter diminishes the 
compressive stress at B, and increases the tensile stress at A. 

The tensile stress per square 
inch at A therefore equals /+/i ; 






P 



Fig. 252. 



Fig. 251. 



where / equals the unit fiber 
b stress due to flexure at A, and/i 
equals the unit tensile stress due 

Plc At P 
— , and/i=- 

in which c = the distance from 
the gravity axis to the outer fiber 



to Pi. Now f=~i L , and f x =— ; 



= AH, and i" = the moment of inertia of the section about H, and 
A = area of the cross-section AB. 

Were the vertical bounding walls at A and B curved surfaces, 
the locus of neutral axes (centroid) through H would be a curve. 
In this case according to E. S. Andrews: * 

At inner edge of section, f b = P[i + (Kldi -f- k 2 )] — A . 
At outer edge, f c = P[Cld 2 -i-k 2 )]-r-A. 

di = distance from centroid to inner edge, inches ; 
d 2 = distance from centroid to outer edge, inches; 

k = radius of gyration of section about centroid in inch units ; 
A = cross-sectional area, square inches ; 
K= (7.845+0.42)- (85-3); 

C= (s -0.16) -r- (5 + 0.2) ; 

s = radius of curvature of centroid divided by (d\ +^2). 
244. Problem. — Let it be required to design the frame of a 
machine to punch f-inch holes in ^-inch steel plates, 18 inches 



American Machinist, Sept. 5, 191 2. 



MACHINE FRAMES. 443 

from the edge. The surface resisting the shearing action of the 
punch = -X-J-"X V' = i.iS square inch. The ultimate shearing 
strength of the material is, say, 50,000 lbs. per square inch. The 
total force P, which must be resisted by the punch frame = 50,000 
X 1.18 = 59,000 lbs. 

The material and form for the frame must first be selected. 
The form is such that forged material is excluded, and difficulties 
of casting and high cost exclude steel casting. The material, 
therefore, must be cast iron. Often the same pattern is used 
both for the frame of a punch and shear. In the latter case, 
when the shear blade begins and ends its cut, the force is not 
applied in the middle plane of the frame, but considerably to one 
side, and a torsional stress results in the frame. Combined torsion 
and flexure are best resisted by members of box form. The frame 
will therefore be made of cast iron and of box section. The dimen- 
sion AB may be assumed so that its proportion to the " reach" 
of the punch appears right ; the width and thickness of the cross- 
section ir ay also be assumed. From these data the maximum 
stress in the outer fiber may be determined. If this is a safe 
value for the material used the design will be right. 

Let the assumed dimensions be as shown in Fig. 252. Then 

A =bid\ — b 2 d 2 = yS square inches. 

7 _ Mi 3 -M 2 3 
12 
= 3002 bi-quadratic inches. 

c=d 1 -H2=c/ / ; /= the reach of the punch + c = 27"; P =59,000 lbs., 
as determined above. Then 

P 59000 
ll= A = ~W~ = 7S ' 

. Pic SQ000X27X9 

=~r = — z =4776, 

I 3002 ' ' 

/x -f/ = 5532 = maximum stress in the section. 



444 



MACHINE DESIGN. 



The average strength of cast iron such as is used for machinery 
castings, is about 20,000 lbs. per square inch. The factor of 
safety in the case assumed equals 20,000-^5532=3.65. This is 
too small. There are two reasons why a large factor of safety 
should be used in this design: I. When the punch goes through 
the plate the yielding is sudden and a severe stress results. This 
stress has to be sustained by the frame, which for other reasons 
is made of unresilient material. II. Since the frame is of cast 
iron, there will necessarily be shrinkage stresses which the frame 
must sustain in addition to the stress due to external force's. 
These shrinkage stresses cannot be calculated and therefore can 
only be provided against by a large factor of safety. 

Cast iron is strong to resist compression and weak to resist 
tension, and the maximum fiber stress is tension on the inner 
side. The metal can therefore be more satisfactorily distributed 
than in the assumed section, by being thickened where it sustains 
tension, as at A, Fig. 253. If, how- 
ever, there is a very thick body 
of metal at a, sponginess and ex- 
cessive shrinkage would result. 
The form B would be better, the 
metal being arranged for proper 
cooling and for the resisting of 
flexure stress. 





Fig. 253. 



Fig. 254. 



Dimensions may be assigned to a section like B and the 
cross-section may be checked for strength as before. See Fig. 254. 
GG, a line through the center of gravity of the section, is found to 
be at a distance of 7.05 inches from the tension side.* The 



* A simple and satisfactory method for obtaining a close approximation to 



MACHINE FRAMES. 445 

required values are as follows: £ = 7.05 inches; / = reach of 
punch + <: = 18 + 7.05 =25.05 inches; .4=156.25 square inches; 
7=5032.5 bi-quadratic inches; P = 59,000 lbs. 

Then /, -4- 59 ^ =377-6 lbs.; 

A 156.25 3U 

Pic 5Q000X25.05X7.05 
/ = -p = — = 2070.4 lbs. 



/i+/ = 2448 lbs. = maximum fiber stress in the section. The 
factor of safety =20,000-^2448=8.17.* This section, therefore, 
fulfills the requirement for strength, and the material is well 
arranged for cooling with little shrinkage, and without spongy 
spots. The gravity axis may be located, and the value of I deter- 
mined by graphic methods. See Hoskins's "Graphic Statics.'' f 
Let the section CD, Fig. 250, be considered. Fig. 255 shows 
the part at the left of CD free. K is the projection of the gravity 
axis of the section. As before, put in two opposite forces, Pa 
and P 4 , equal to each other and to P, and having their common 
line of action parallel to that of P, at a distance /] from it. P 
and P 4 now form a couple, whose moment =P/i, tending to produce 
flexure about K. P 3 must be resolved into two components, 
one P 3 7, at right angles to the section considered, tending to 
produce tensile stress; and the other JK, parallel to the section, 
tending to produce shearing stress. The greatest unit tensile 

the true gravity axis of irregular figures is as follows: On a piece of thin but uni- 
form cardboard lay out the figure to scale. Cut it out carefully with a sharp 
knife. Balance the figure exactly, by trial, on a knife-edge. The line of contact 
with the knife-edge is the gravity axis. Its position may be marked and its loca- 
tion measured to scale. 

* This discussion neglects the action of gravitv which would exert a counter- 
balancing moment, reducing the maximum tensile fiber stress below the value 
found. This makes the actual factor of safety greater than the apparent factor 
of safety. 

t The student will be familiar with analytical methods for their determina- 
tion from his study of the ''Mechanics of Materials." 



446 



MACHINE DESIGN. 



stress in this section will equal the sum of that due to flexure 
and that due to tension 



The greatest unit shear =/ s 



JK 
A ' 



In the section PE, Fig. 250, which is parallel to the line of 
action of P, equal and opposite forces, each = P, may be intro- 
duced, as P 5 and P 6 . P and P 6 will then form a couple with an 




>////y///////////////^^^^ 

Fig. 256. 



arm / 2 , and P 5 will be wholly applied to produce shearing stress. 
The maximum unit tensile stress in this section will be that due 
to flexure, f=Pl2C--I, and the maximum unit shear will be 
j 8 =P-r-A. Any section may be thus checked. 



MACHINE FRAMES. 



447 



The dimensions of several sections being found, the outline 
curve bounding them should be drawn carefully, to give good 
appearance. The necessary modifications of the frame to pro- 
vide for support, and for the constrainment of the actuating 
mechanism, may be worked out as in Fig. 256. A is the pinion 
on the pulley shaft from which the power is received; B is the 
gear on the main shaft ; C, D, and G are parts of the frame added 
to supply bearings for the shafts; E furnishes the guiding surfaces 
for the punch "slide." The method of supporting the frame is 
shown, the support being cut under at F for convenience to the 
workman. The parts C, D, E, and G can only be located after 
the mechanism train has been designed. 




Fig. 257. 

245, Slotting-machine Frame. — See Fig. 257. It is specified 
that the slotter shall cut at a certain distance from the edge of 
any piece, and the dimension AH is thus determined. The 
table G must be held at a convenient height above the floor, and 
RK must provide for the required range of "feed." K is cut 
under for convenience of the workman, and carried to the floor 
line as shown. It is required to "slot " a piece of given vertical 
dimension, and the distance from the surface of the table to E 
is thus determined. Let the dimension LM be assumed so that 



44§ MACHINE DESIGN. 

it shall be in proper proportion to the necessary length and height 
of the machine. The curves LS and M T may be drawn for 
bounding lines of a box frame to support the mechanism. M 
should be carried to the floor line as shown, and not cut under. 
None of the part DNE, nor that which serves to support the 
cone and gears on the other side of the frame, should be made 
flush with the surface LSTM, because nothing should interfere 
with the continuity of the curves LS and TM. The supporting 
frame oj a machine should be clearly outlined, and other parts should 
appear as attachments. The member VW should be designed so 
that its inner outline is nearly parallel to the outline of the cone 
pulley, and should be joined to the main frame by a curve. The 
outer outline should be such that the width of the member increases 
slightly from W to V, and should also be joined to the main 
frame by a curved outline. In any cross-section of the frame, 
as XX, the amount of metal and its arrangement may be con- 
trolled by the core. It is dictated by the maximum force, P, 
which the tool can be required to sustain. The tool is carried by 
the slider of a slider-crank chain. Its velocity varies, therefore, 
from a maximum near mid-stroke, to zero at the upper and lower 
ends of its stroke. The belt which actuates the mechanism runs 
on one side of the steps of the cone pulley, at a constant velocity. 
Suppose that the tool is set (accidentally) so that it strikes the 
table just before the slider has reached the lower end of its stroke. 
The resistance, R, offered by the tool to being stopped, multiplied 
by its (very small) velocity, equals the difference of belt tension 
multiplied by the belt velocity (friction and inertia neglected).* 
R, therefore, would vary inversely as the slider velocity, and hence 
may be very great. Its maximum value is indeterminate. A 
" breaking piece " may be put in between the tool and the crank, 
Then when R reaches a certain value, the breaking piece fails. 

* See Chapter V. 



MACHINE FRAMES. 



449 



The stress in the stress-members of the machine is thereby limited 
to a certain definite value. From this value the frame may be 
designed. Let P = upward force against the tool when the break- 
ing piece fails.* Let / = the horizontal distance from the line of 
action P to the gravity axis of the section XX. Then the section 
XX sustains flexure stress caused by the moment PI, and tensile 
stress equal to P. The maximum unit stress in the section 



= / + /! = 



Pic 



A' 



A section may be assumed and checked for safety, as for the 
punching-machine in § 244. 

246. Stresses in the Frame of a Side-crank Steam-engine. — 

Fig. 258 is a sketch in plan of a side -crank engine of the " girder 




Fig. 258. 

bed " type. The supports are under the cylinder C, the main 
bearing E, and the out-board bearing D. A force P is applied 
in the center line of the cylinder, and acts alternately toward the 
right and toward the left. In the first case it tends to separate 
the cylinder and main shaft; and in the second case it tends to 
bring them nearer together. The frame resists these tendencies 
with resulting internal stresses. 

Let the stresses in the section AB be considered. The end of 
ths frame is shown enlarged in Fig. 259. If the pressure from the 
piston is toward the right, the stresses in AB will be: I. Flexure 

* P is limited to the friction due to screwing up the four bolts which hold the 
tool. 



45o 



MACHINE DESIGN. 



due to the moment PI, resulting in tensile stress below the gravity 
axis, N, with a maximum value at b, and a compressive stress 
above N with a maximum value at a. II. A direct tensile stress, 
= P, distributed over the entire section, resulting in a unit stress = 
P + A=fi lbs. per square inch. This is shown graphically at n. 
Fig. 259. #i&i is a datum line whose length equals ab. Tensions 
are laid off toward the right and compressions toward the left. 



k\e\a\ 

3 



N f 



L 



61 C]d\ 
(n) 



a 


zeilci A 


'I 




,,| N 


// 


X ! 


d^c 2 
(r 


62 

B 


\ j 1 

i 
p i 



Fig. 259. 



The stress due to flexure varies directly as the distance from the 
neutral axis Ni, being zero at iVi. If, therefore, btfi represents 
the tensile stress in the outer fiber, then C1&1 drawn through Ni 
will be the locus of the ends of horizontal lines, drawn through 
all points of ajfi, representing the intensity of stress in all parts 
of the section, due to flexure. If cidi represent the unit stress 
due to direct tension, then, since this is the same in all parts of 
the section, it will be represented by the horizontal distance be- 
tween the parallel lines C1&1 and diei. This uniform tension 
increases the tension biCi due to flexure, causing it to become bidi ; 
and reduces the compression &i#i, causing it to become e x (i\. 
The maximum stress in the section is therefore tensile stress in 
the lower outer fiber, and is equal to b\d\. 

When the force P is reversed, acting toward the left, the 
stresses in the section are as shown at m, Fig. 259: compression 
due to flexure in the lower outer fiber equal to c 2 b2', tension due 
to flexure in the upper outer fiber equal to a 2 k 2 ; and uniform 
compression over the entire surface equal to d 2 c 2 - This latter 
increases the compression in the lower outer fiber from b 2 c 2 to 
b 2 d 2 , and decreases the tension in the upper outer fiber from a 2 k 2 



MACHINE FRAMES. 



451 



to a 2 e2- The maximum stress in the section is therefore compres- 
sion in the lower outer fiber equal to b 2 d 2 . The maximum stress, 
therefore, is always in the side of the frame next to the connecting- 
rod. 

If the gravity axis of the cross-section be moved toward the 
connecting-rod, the stress in the upper outer fiber will be increased, 
and that in the lower outer fiber will be proportionately decreased. 
The gravity axis may be moved toward the connecting-rod by 
increasing the amount of material in the lower part of the cross - 
section and decreasing it in the upper part. 

The stress in any other section nearer the cylinder will be due 
to the same force, P, as before; but the moment tending to pro- 
duce flexure will be less, because the lever arm of the moment is 
less and the force constant. 

247. Heavy-duty Engine Frame. — Suppose the engine frame 
to be of the type which is continuous with the supporting part as 
shown in Fig. 260. Let Fig. 261 be a cross-section, say at AB. 
O is the center of the cylinder. The force P is applied at this 




Fig. 261. 



point perpendicular to the paper. C is the center of gravity of 
the section (the intersection of two gravity axes perpendicular 
to each other, found graphically). Join C and O, and through 



452 MACHINE DESIGN. 

C draw XX perpendicular to CO. Then XX is the gravity 
axis about which flexure will occur.* The dangerous stress will 
be at F, and the value of c will be the perpendicular distance from 
F to XX. The moment of inertia of the cross-section about 
XX may be found, =7; /, the lever-arm of P, = OC. The 
stress at F, / + /i must be of safe value. 



Pic 


in known terms. 


p 


in known terms. 


1 area of section' 



248. Closed Frames. — Fig. 262 shows a closed frame. The 
members G and H are bolted rigidly to a cylinder C at the top, and 
to a bedplate, DD, at the bottom. A force P may act in the center 
line, either to separate D and C, or to bring them nearer together. 
The problem is to design G, H, and D for strength. If the three 
members were "pin connected" (see Fig. 263), the reactions of 
C upon A and B at the pins would act in the lines EF and GH. 
Then if P acts to bring D and C nearer together, compression 
results in A, the line of action being EF; compression results 
in B, the line of action being GH. These compressions being in 
equilibrium with the force P, their magnitude may be found by 
the triangle of forces. From these values A and B may be designed. 
C is equivalent to a beam whose length is /, supported at both 

* This is not strictly true. If OC is a diameter of the "ellipse of inertia," 
flexure will occur about its conjugate diameter. If the section of the engine frame 
is symmetrical with respect to a vertical axis, OC is vertical, and its conjugate 
diameter XX is horizontal. Flexure would occur about XX, and the angle be- 
tween OC and XX would equal 90 . As the section departs from symmetry 
about a vertical, XX, at right angles to OC, departs from OC's conjugate, and 
hence does not represent the axis about which flexure occurs. In sections like 
Fig. 259, the error from making ^=90° is unimportant. When the departure 
from symmetry is very great, however, OC's conjugate should be located and 
used as the axis about which flexure occurs. For method of drawing " ellipse of 
inertia" see Hoskins's " Graphic Statics." 



MACHINE FRAMES 



453 



ends, sustaining a transverse load P, and tension equal to the 
horizontal component of the compression in A or B. The data 
for its design would therefore be available. Reversing the direc- 
tion of P reverses the stresses; the compression in A and B becomes 
tension; the flexure moment tends to bend C convex downward 
instead of upward, and the tension in C becomes compression. 
But when the members are bolted rigidly together, as in 



Fig. 262. 




Fig. 263. 

Fig. 262, the lines of the reactions are indeterminate. Assump- 
tions must therefore be made. Suppose that G is attached to D 
by bolts at E and A. Suppose the bolts to have worked slightly 
loose, and that P tends to bring C and D nearer together. There 
would be a tendency, if the frame yields at all, to relieve pressure 
at E and to concentrate it at ^4. The line of the reaction would 
pass through A and might be assumed to be perpendicular to 



454 MACHINE DESIGN. 

the surface AE. Suppose that P is reversed and that the bolts 
at A are loosened, while those at E are tight. The line of the 
reaction would pass through E, and might be assumed to be 
perpendicular to EA. MN is therefore the assumed line of the 
reaction, and the intensity R=P-^2. In any section of G, as 
XX, let K r be the projection of the gravity axis. Introduce at 
K' two equal and opposite forces equal to R and with their lines 
of action parallel to that of R. Then in the section there is flex- 
ure stress due to the flexure moment Rl, and tensile stress due to 
the component of R 2 perpendicular to the section, =Rs. Then 
the maximum stress in the section =/ + /i- 

1- A \ h- j- 

A section may be assumed, and A, I, and c become known; 
the maximum stress also becomes known, and may be compared 
with the ultimate strength of the material used. 

Obviously this resulting maximum stress is greater when the 
line of the reaction is MN than when it is KL. Also it is greater 
when MN is perpendicular to EA than if it were inclined more 
toward the center line of the frame. The assumptions therefore 
give safety. If the force P could only act downward, as in a 
steam hammer, KL would be used as the line of the reaction. 

The part D in the bolted frame is not equivalent to a beam 
with end supports and a central load like C, Fig. 263, but more 
nearly a beam built in at the ends with central load, and it may 
be so considered, letting the length of the beam equal the hori- 
zontal distance from E to F, =l x . Then the stress in the mid- 

Ph 

section will be due to the flexure moment —5-, and the maximum 

o 

PI c 
stress = / = sr~- The values c and 7 may be found for an assumed 
01 

section, and / becomes known. 






MACHINE FRAMES. 455 

249. Steam-hammer Frames. — Steam-hammers are made both 
with " open-side " and "closed " frames. They may therefore be 
designed by methods already given, if the maximum force applied 
is known. The problem is, therefore, to find the value of this 
maximum force. 

There are two types of steam-hammers: 

Type 1. Single-acting. A heavy hammer-head attached to 
a steam-piston is raised to a certain height by steam admitted 
under the piston. The steam is then exhausted and the hammer- 
head with attached parts falls by gravity to its original position. 
The energy of the blow = Wl, where W is the falling weight and 
/ is the height of fall. 

Type 2. Double-acting. A lighter hammer-head is lifted by 
steam acting under its attached piston, and during its fall steam 
is admitted above the piston to help gravity to force it downward. 
The energy of the blow-=W7 (as before) plus the energy received 
from the expansion of the steam; or, if the steam acts throughout 
the entire stroke, the energy of blow = Wl + pAl, where p is the 
mean pressure per square inch and A is the area of the upper 
side of the piston. 

250. Stresses in Single-acting Frames. — In type 1, when the 
action is as described, a force acts downward upon the frame during 
the lifting of the hammer. The intensity of this force =pA =the 
mean pressure of steam admitted multiplied by area of piston, 
and the line of action is the axis of the piston-rod. During the 
fall of the hammer the cylinder and frame act simply as a guide, 
and no force is applied to the frame except such as may result 
from frictional resistance. The hammer strikes an anvil which 
is not attached to the frame, but rests upon a separate foundation. 

But a greater force than pA may be applied to the frame. 
In order that a cushioned blow may be struck, the design is such 
that steam may be introduced under the piston at any time during 
its downward movement, and this steam is compressed by the 



456 



MACHINE DESIGN. 



advancing piston. A part of the energy of the falling hammer 
is used for this compression. The pressure in the cylinder result- 
ing from this compression is communicated to the lower cylinder- 
head and through it to the frame. Under certain conditions steam 
might be admitted at such a point of the stroke that all of the 
energy of the falling hammer might be used in compressing the 
steam to the end of the stroke. The hammer would then just 
reach the anvil, but would not strike a blow. 

Fig. 264, a shows by diagram a hammer of type 1. Steam is 
admitted, the piston is raised, the exhaust- valve is opened, and 




Fig. 264. 



the piston falls. But at some point in the stroke steam is again 
admitted, filling the cylinder, and the valve is closed. Com- 
pression occurs and absorbs all or part of the energy Wl. In the 
latter case the hammer will strike the anvil a blow whose energy 
is equal to Wl minus the work of compressing the steam in C. 



MACHINE FRAMES. 457 

The compression is shown upon a pressure-volume diagram, 
Fig. 264, b. Progress along the vertical axis from B toward E 
corresponds to the downward movement of the hammer. Vertical 
ordinates therefore represent space, 5, moved through by the 
hammer; or, since SA = volume displaced by the piston, the 
vertical ordinates may also represent volumes. EF represents 

V 2 

the volume V2 of the clearance space, or -7-, the piston movement 

which corresponds to the clearance. Horizontal ordinates meas- 
ured from BF represent absolute pressures per square inch. 
Let pi represent the absolute boiler pressure represented by DK. 
TN is the line of atmospheric pressure. During the lifting of the 
hammer the upper surface of the piston is exposed to atmospheric 
pressure and the lower surface is exposed to pressure just suffi- 

W 
cient to raise the hammer, =-?-. The work of lifting is represented 

by the area NTHG. This work equals the energy, Wl, which 
the hammer must give out in some way before reaching the anvil 
again. When the piston has fallen to some point, as D, steam 
may be let in below it at boiler pressure, DK. The advancing 
piston will compress this steam, and KM will be the compression 
curve.* The work of compression is represented by the area 
RKMN. If the compression is to absorb all the energy Wl, the 
area which represents the work of compression must equal the 
area which represents Wl. Hence area NTHG must equal 
RKMN: or, since the area RLGN is common to both, the area 
RTHL must equal the area LKMG. The point at which com- 
pression must begin in order to cause this equality may be found 
by trial. The greatest unit pressure reached by compression is 
represented by EM. The greatest pressure, p 2 , upon the lower 
cylinder-head is represented by NM, since atmospheric pressure 

* Assuming />K =consiant. 



45^ MACHINE DESIGN. 

acts on the outside. The corresponding total force communicated 
to the frame = p 2 A=P* 

If compression had begun earlier the energy would have been 
absorbed before the hammer reached the anvil, the piston would 
have stopped short of the end of the stroke, the compression curve 
would have been incomplete, and the greatest pressure would have 
been less than EM. Obviously if compression had begun later 
the greatest pressure would have been less than EM. Therefore 
the force P, = p2A, with the cylinder's axis for its line of action, 
Is the greatest force that can be applied to the frame in the regular 
working of the hammer. 

A greater force might be accidentally applied. For, suppose 
that water is introduced into the cylinder in such quantity that 
the piston reaches it before the hammer reaches the anvil, then 
all the energy will be given out to overcome the resistance of the 
water. The resulting force is indeterminate, because the space 
through which the resistance acts is unknown. This force may 
be very great. The force applied to the frame may be limited 
by the use of a "breaking- piece." Thus the studs which hold 
on the lower cylinder-head may be drilled j so that they will 
break under a force KP, in which K is a factor ot safety and P 
is the force found above. Then the breaking-piece will be safe 
under the maximum working force, but will yield when an acci- 
dental force equals KP, thus limiting its value. The frame may 
be designed for a maximum force KP. 

251. Problem, Type 1. — Let W, weight of hammer and 

attached parts, =2000 lbs.; /, maximum length of stroke, =24 

inches ; A , effective area of piston, — 50 square inches ; clearance 

= 15 pei cent; boiler pressure =85 lbs. by gauge. Steam is 

admitted to lift the hammer, pressure being controlled by throttling. 

* la which A is the effective area of the piston, i.e., area of the piston less 
;area ot the rod, 
t See page 156 



MACHINE FRAMES. 



459 



The pressure per square inch that will just lift the hammer = 
2000 lbs. ■*- 50 square inches =40 lbs. In Fig. 264, NG repre- 
sents 40 lbs., and NT represents the volume displaced by the 
piston during a complete stroke. Hence NTHG represents the 
work of lifting the hammer, or the energy that must be absorbed 
just as the hammer reaches the anvil. Trial shows that to 
accomplish this, compression must begin at just about 6 inches 
from the end of the stroke. The maximum resulting pressure, 
represented by NM, equals 258 lbs. per square inch. The total 
pressure acting downward on the frame =p2& = 258 X 50 = 12,900 
lbs. = P. If the factor of safety, K, is 5, the strength of the 
breaking-piece =KP = 5X12,900=64,500 lbs. This is the maxi- 
mum force, and hence may be used as a basis of the frame design. 

252. Stresses in Double-acting Frames. — In type 2 the 
maximum working force may be found by a similar method. In 
Fig. 265, NG represents the pressure per square inch of piston 
necessary to raise the hammer. The area NTHG represents 
the energy stored in the hammer by lifting. The area HSJL 
represents the work done by steam at boiler pressure acting on 
the upper piston face while the piston descends to D. At this 
point steam is exhausted above the piston and let in below it, and 
compression takes place during the remainder of the stroke. To 
absorb ail the energy of the hammer by compression, the areas 
NTSJLG and RKMN must be equal. The area NRLG is 
common to both; hence the area LKMG must equal the area 
RTSJ. The point at which compression must begin in order 
to cause this equality may be found by trial. 

253. Problem, Type 2. — Let W, weight of hammer and 
attached parts, =6co lbs.; /, maximum length of stroke, =24 
inches; A, effective area of piston (both faces), =50 square 
inches ; clearance = 15 per cent; boiler pressure = 85 lbs. by gauge. 
The construction in Fig. 265 shows that compression, beginning 
at C)\ inches before the end of the piston's stroke, absorbs all 



460 



MACHINE DESIGN. 



the energy of the hammer, and gives 325 lbs. as a maximum 
pressure per square inch. Then the maximum working force 
= 3 2 5 X 50 = 16,250. If K = 5, the strength of the breaking-piece 
= 16,250 X 5 =81,250 lbs. 



K 






BTH 





R 


L K 


J 




N 


G 


■ _-_JV> 







CD. 



Fig. 265. 

254. Other Stresses in Hammer Frames. — An accidental force 
acting upward may be applied to the hammer frame. The boiler 
pressure is necessarily greater than that which is necessary to 
lift the hammer.* Thus in § 251 a pressure of 40 lbs. per square 
inch is sufficient to lift the hammer, but the boiler pressure is 
85 lbs. per square inch. If the throttle- valve were opened wide 
and held open during the movement of the hammer upward, the 
energy stored in the hammer when it reaches its upper position 
would equal the product of boiler pressure, piston area, and length 
of stroke, =85X50X24 = 106,000 in.-lbs. The energy necessary 



* So that it may be possible to work the hammer when the steam-pressure is 
lower in the boiler. 



MACHINE FRAMES. 461 

to just lift the hammer is 40X50X24=48,000 in. -lbs. The 
difference between these two amounts of energy, =58,000 in. -lbs., 
will exist as kinetic energy of the moving hammer; and it must 
be absorbed before the hammer can be brought to rest in its 
upper position. The force which would result from stopping 
the hammer would be dependent upon the space through which 
the motion of the hammer is resisted. Springs are often provided 
to resist the motion of the hammer when near its upper position. 
These springs increase the space factor of the energy to be given 
out and thereby reduce the resulting force. An automatic 
device for closing the throttle-valve before the end of the stroke 
and introducing steam for compression above the piston may 
be used. The steam is then a fluid spring. 

255. Design of Crane Frames. — A crane frame is to be de- 
signed from the following specifications : Maximum load, 5 tons 
= 10,000 lbs.; radius = maximum distance from the line of lifting 
to the axis of the mast, =18 feet; heighc of mast = 20 feet; radial 
travel of hook in its highest position = 5 feet; axis of jib to be 15 
feet above floor line. Fig. 266 shows the crane indicated by the 
center lines of its members. 

The external forces acting on the crane may be considered first. 
A load of 10,000 lbs. acts downward in the line ab. This is held 
in equilibrium by three reactions : one acting horizontally toward 
the left through the upper support, i.e., along the line be, another 
acting horizontally toward the right through the lower support, 
i.e., in the line ad; a third acting vertically upward at the lower 
support, i.e., in the line cd. The crane is a "four-force piece." 
One force, AB, is completely known, the other three are known 
only in line of action. Produce ab and be to their intersection at 
M. The line of action of the resultant of ab and be must pass 
through M. The resultant of cd and da must be equal and 
opposite to the resultant of ab and be, and must have the same 
line of action. But the line of action of the resultant of cd and 



462 



MACHINE DESIGN. 



da must pass through N. Hence MN is the common line of 
action of the resultants of ab and be and of cd and da. Draw 
the vertical line AB * representing 10,000 lbs. upon some assumed 
scale ; from B draw BC parallel to be, and from A draw AC parallel 
to MN. The intersection of these two lines locates C and deter- 
mines the magnitude of BC. Now AC is the resultant of AB 
and BC y and CA, equal and opposite, is the resultant of CD 




Force Diagram 



Fig. 266. 

and DA. Therefore CA has but to be resolved into vertical and 
horizontal components to determine the magnitudes of CD and 
DA. The force polygon is therefore a rectangle and CD=AB f 
smdBC=DA. 

From the forces AD and CD acting at N, the supporting 
journal and bearing at the base of the crane may be designed; 
and from the force BC, acting at V, the upper journal and bearing 
may be designed. 

256. Jib. — The forces acting on the jib are, first, AB acting 
vertically downward at its end; second, an upward reaction 

*See Force Diagram, Fig. 266. 



MACHINE FRAMES. 



463 



from the brace at H, whose line of action coincides with the axis 
of the brace ; * third, a downward reaction at L where the jib 
joins the mast, whose line of action must coincide with the line 
of action of the resultant of AB and the brace reaction. LK is 
therefore this line of action.. 

Fig. 267. 




In the force diagram draw BE parallel to the center line of the 
brace, and draw EA parallel to LK. Then BE will represent the 
brace reaction, and EA will represent the reaction at L in the line 
ae. Let RQ, Fig. 267, represent the jib isolated, ae, be, and ab 
are the lines of action of the three forces acting upon it. The 
vertical components of these forces are in equilibrium, and tend 
to produce flexure in the jib. The horizontal components are 
in equilibrium and tend to produce tension in the jib. The vertical 



* Considering the joint between the brace and jib equivalent to a pin con- 
nection. 



464 MACHINE DESIGN. 

force acting at R is FA* =5000 lbs., the vertical force acting at 
T is FB, =15,000 lbs., and the vertical force acting at Q is AB y 
= 10,000 lbs. 

Flexure is also produced in the jib by its own weight acting as 
a uniformly distributed load. 

In order to design the jib, standard rolled forms may be 
selected which will afford convenient support for the sheave car- 
riage. Two channels located as shown in Fig. 268 will serve for 
this crane. For trial 12-inch heavy channels are chosen. From 
Carnegie's Hand-book, the moment of inertia for each channel 
about an axis perpendicular to the web at the center = 7 = 248; 
r = 6 inches; the weight, w, of two channels per inch of length 
= 8J lbs. 

The total weight of the two channels =wl = 8JX 18 X 12 = 

1800 lbs. The vertical reaction, P u at R, Fig. 267, due to this 

(wl^ wl\ 2 \ 
weight is P\ = ( — ■ ) -5- h, from the equation of moments, 

due to the weight of jib about the point T. Introducing numerical 

values, Pi =450 lbs. The total reaction at R is therefore 5000 

— 450=4550 lbs. A diagram of moments of flexure may now be 

drawn under the jib, Fig. 267. Considering the portion TQ, the 

wli 2 
moment at T = Pli + =741,600 in. -lbs. Divide TQ into 

three equal parts. At the division nearest T the moment = 

w/3 2 
P/ 3 + ; in which / 3 = the distance from Q to the section con- 
sidered. The moment at the other two points may be found 
by similar method. 

<7£l/ 2 

The moment at any point at the left of P = 4550X^4 + — — ; 

in w T hich / 4 is the distance from R to the section considered. From 
the values thus found the diagram of flexure moments may be 

* See force diagram, Fig. 266. 



MACHINE FRAMES. 



465 



drawn. The maximum value is at T, where 3f 6 = 741,600 in. -lbs. 
The resulting fiber stress = / =-7— =- — o = 8971 lbs. The 

horizontal component acting at R is equal to FE (see force dia- 
gram, Fig. 266) =12,000 lbs. An equal and opposite horizontal 
force must act at T. Between T and Q there is no tensile stress 
due to the forces AB, BE, and AE. 

Fig. 269. 




Fig. 270. 



Another force which modifies this result needs to be considered. 
Let AB, Fig. 269, be the upper surface of the jib. The load is 
supported as shown. The chain which is fastened at B passes 
over the right-hand carriage sheave, down and under the hook 
sheave, up and over the left-hand carriage sheave, horizontally 
to the sheave at A, and thence to the winding drum. If a load 
of 10,000 lbs. is supported by the hook there will be, neglecting 
friction, a tension of 5000 lbs. throughout the entire length of the 



466 MACHINE DESIGN. 

chain from B to the winding drum. There is therefore a force 
of 5000 lbs. tending to bring A and B nearer together, and hence 
to produce compression in the jib* The resultant tension be- 
tween R and T is 12,000-5000 = 7000 lbs., while between T and 
Q there is a compression of 5000 lbs. The cross-sectional area 
of the two channels selected =30 square inches. Hence the unit 
tensile stress = /i = 7000-^30 = 233 lbs. The maximum unit ten- 
sile stress in the jib =/ + /i =8971 +233 =9204 lbs. per square inch. 

If the channels are of steel, their unit tensile strength will 
probably equal 60,000 lbs. per square inch. The factor of 
safety = 60,000-^9204=6.5. In a crane a load may drop through 
a certain space by reason of the slipping of a link that has been 
caught up, or the failure of the support under the load while the 
chain is slack. When this occurs a blow is sustained by the stress 
members of the crane. The energy of this blow equals the load 
multiplied by the height of fall. But the stress members of the 
crane are long, and the .yielding is large. Hence the space through 
which the blow is resisted is large and the resulting force is less 
than with small yielding. In other words, the stress members act 
as a spring, reducing the force due to shock. Hence, in a crane 
of this type, the ductile and resilient material is liable to modified 
shock, and a factor of safety =6.5 is large enough. 

The jib might also be checked for shear, but in general it will 
be found to have large excess of strength. 

257. Mast. — Fig. 270 shows the mast by its center line with 
the lines of action of the forces acting upon it. It is equivalent 
to a beam supported at C and D with a load at A. The moment 
of flexure at A equals the force acting in the line bc\ multiplied 
by the distance CA in inches, =9000 lbs. X 60 inches = 540,000 



* There is also flexure due to this force multiplied by the distance from the 
centre line of the horizontal chain to the gravity axis of the jib. This is small 
and may be neglected. 

t The force BC in Fig. 266. 



MACHINE FRAMES. 



467 



in. -lbs. = M h . The flexure moment is a maximum at this point. 

and decreases uniformly toward both ends. The moment diagram 

is therefore as shown. The maximum fiber stress due to this 

flexure moment =f =M b c^L Selecting two light 12-inch channels, 

* • 1 - S4ooooX6 

7 = 176; c = 6 inches; / = ^-=0200 lbs. 

1 2X176 

The tension in the mast equals the vertical component of the 
force acting in the line ae* = $ooo lbs. (Actually reduced to 
4550 by the effect of the weight of the jib.) The compression in 
the mast due to the tension in the chain = 5000 lbs. between A 
and the point of support of the winding drum B. The tension 
and compression therefore neutralize each other, except below B y 
where the flexure moment is small. Hence the maximum unit 
stress in the mast is 9200 lbs. The factor of safety = 60000 ^-9200 
= 6.5, which is safe as before. This also may be checked for shear. 

258. Brace. — The compression stress in the brace is 19,000 
lbs.,f and the length, 19 feet, is such that is needs to be treated 
as a "long column." Because of the yielding of joints and of 
the other stress members, the brace is intermediate between a 
member with "hinged ends" and "flat ends"; therefore for safety 
it should be considered as hinged. In the treatment of long 
columns, the "straight-line formula" will be used.! This 
formula is of the form 

A r r 

P is the total force that will cause incipient buckling, and hence 
the force that will destroy the column; A is the cross-sectional 
area of the column; p is the unit stress that will cause buckling; 



* The force AE in Fig. 266. 

I See force diagram, Fig. 266. 

X For discussion of long-column formulae see "Theory and Practice of Mod- 
ern Framed Structures," by Johnson, Bryan, and Turneaure, page 143. Published 
by John Wiley & Sons. 



468 MACHINE DESIGN. 

B and C are constants derived from experiments on long columns 
(the values of B and C vary with the method of attachment of the 
ends of the column, and with the material of the column); / is 
the length of the column in inches ; and r is the radius of gyration 
of the cross-section, =\/l~A, I being the moment of inertia of 
the cross-section referred to the axis about which buckling takes 
place. 

Values of B and C are as follows: 

P I 

For wrought iron, hinged ends, -j =42000 — 157—. 



( c i i 1 c 



P I 

flat " -7=42000 — 1 28-. 



P I 

11 mild steel, hinged " -7=52500 — 220-. 

JL Y 

11 " " nat " 5 = 52500-1797 



The brace will be of mild steel channels, and the ends will be 
considered as hinged. The formula to be used is therefore 

P I 

-^•=52500-220-, 



from which 



P = (52500 — 220- J A, 



Channel bars may be selected and values of r and A become 
known from tables. For trial 5-inch light channels are chosen. 
Carnegie's tables give for 2 channels, r=i.g$ inches, and A = 
3.9 square inches. Introducing these values in the above equation, 
with /=io/Xi2 = 228 // , gives P= 106,970 lbs. Since the maxi- 
mum compression force sustained by the brace =19,000 lbs., 



MACHINE FRAMES. 



469 



the factor of safety = 106,970-7-19,000 = 5.6 + . This is a smaller 
value than those for the jib and mast, but it is probably inadvisable 
to use larger channels because of convenience in making the 
connection with the jib and mast. 

But the brace must be made safe against side buckling. The 
two channels may be considered as acting as a single member if 
they are braced laterally. The lateral bracing will be determined 
later. In Fig. 271 the moment of inertia about the axis X for 



# 



Y A a 

b~ 



jr~ '— -1 




Fig. 271. 



Fig. 272 



each channel = 7 (from table.) If the moment of inertia of each 
about the axis Y be made = 7, the radius of gyration will be the 
same about both axes, the values in the above equation will be 
the same, and there will be the same safety against side buckling 
as against buckling in the plane through the axis of the mast. 
Therefore it is only necessary to make the distance, a, of each 
channel from the axis of Y such that Iy = 7- The moment of 
inertia of one channel about its own gravity axis GG =0.466. Its 
moment of inertia about Y = 7 = I G +Ax 2 . Solving, x 2 = (Iy — Iq) 
s-A, whence 



7-0.466 

c* = 

i-9S 

# = 1.828. 



3-35) 



Hence the distance apart of the gravity axes = 1.828X2 =3.65. 
But the gravity axis is 0.44 inch from the face of the web, i.e., 
x — a =0.44. Therefore the distance, b, between the channels 
= 3.65-0.88 = 2.77 inches. Convenience in construction would 



470 MACHINE DESIGN. 

undoubtedly dictate a greater distance, and hence greater safety 
against side buckling. 

The position of these two channels relative to each other must 
be insured by some such means as diagonal bracing. See Fig. 272. 
The distance, /, must be such that the channels shall not buckle 
separately under half the total load. Solving the long-column 
formula gives 

l-(s*5oo-0^- 

in which P is the load sustained by each channel ( = 190004-2 = 
9500 lbs.) multiplied by the factor of safety, say 6. The radius of 
gyration, r, is about a gravity axis parallel to the web, =Vl -r-A =* 
V0.466 -r- 1. 95 =0.488. 

_ / 9500X6X0.488 
/H52500-— )— -^ — = 51.8. 

\° 1.95 / 220 ° 

The value of /, therefore, must not be greater than 51.8 inches 
but it may be less if convenience, or the use of standard braces 
requires. 

259. Crane Frame with Tension Rods. — The brace in the 
crane just considered may be replaced by tension rods, as shown 
in Fig. 273. This allows the load to be moved radially through- 
out the entire length of the jib. The force polygon, Fig. 274, 
shows tension equal to 37,000 lbs. in the tension rods, and com- 
pression in the jib -=35,800 lbs. If the tension rods are made of 
mild steel with an ultimate tensile strength of 60,000 lbs., and 
a factor of safety =6, the cross-sectional area must equal 
37,000X6-^60,000=3.7 square inches. If two rods are used the 
minimum diameter of each = 1.535 > sav > J & inches. 

The mast is a flexure member 20 feet long supported at the 
ends, and sustaining a transverse force of 35,800 lbs. at a distance 
of 5 feet from the upper end. The upper end reaction is there- 



MACHINE FRAMES. 



471 



fore 35,8ooX-2-|- = 26,850 lbs., and the maximum moment of 
flexure at iV = 26,850X60 = 1,611,000 in.-lbs. Selecting two 

• 1_ 1_ 1 tv, r l6lIOOOX7.5 

1 5-inch heavy eye-beams, 7 = 750X2; c = 7i; .*. / = — 

750X2 

= 8055 lbs. = maximum unit stress in the mast. The factor of 



safety 



60000 



= 7.4, safe. 



Fig. 273. 




Fig. 274. 

The moment of flexure in the jib is a maximum when the 

maximum load is suspended at its center. The maximum flexure 

PI wl 2 
moment due to the load and the weight of the jib =M b = — -f — , 

4 8 

in which P = load = 10,000 lbs. ; w = weight of two channels per 
foot of length, =100 lbs. if 12-inch heavy channels are chosen. 
Substituting numerical valuer M b = 49,050 ft.-lbs. =588,600 in.- 
lbs. The resulting maximum fiber stress due to flexure, / = — ^~ 
588600X6 
2X248 



7120 lbs. 



472 MACHINE DESIGN- 

Compression in jib due to chain tension = 10,000 lbs.; 

Compression in jib due to load =35,800 lbs.; 

Combined compression due to both = 10,000 X 35,800 =45,800 lbs. 

Unit compression = combined compression -4- area of the two 

45800 
channels = = 1 526 lbs. 

Maximum fiber stress due to combined flexure and com- 

00000 
pression = 7120 + 1526 =8647 lbs. The factor of safety = 

0040 

= 6.96. If a smaller factor of safety were desired, smaller channels 

could be used. The jib may be checked for shear. 

The load might be moved nearly up to the mast, hence the 
joint at F must be designed for a total shear of 10,000 lbs. The 
pin and bearing at G, as well as the supporting framework for 
the bearing must be capable of sustaining a lateral force = BC 
=8950 lbs. in any direction. The pivot and step at H must be 
capable of sustaining the lateral force AD = 8g$o lbs., as well as 
a vertical downward thrust of 10,000 lbs. +the weight of the crane. 

260. Pillar-crane Frame. — Fig. 275 shows an outline of the 
frame of a pillar crane. HN represents the floor level; HK 
represents the pillar, which is extended for support to L; KM 
represents one or more tension rods; MH represents the brace. 
The load hangs from M in the line ab. The pillar is supported 
horizontally at H, and vertically and horizontally at L. The 
force polygon shows the horizontal forces at H and L = 30,000 lbs., 
and the vertical force at L= 10,000 lbs. From these the sup- 
ports may be designed. These supports should provide for 
rotary motion of the crane about KL, the axis of the pillar. 
The brace may be treated as in the jib crane, the compressive 
force being 21,400 lbs. The tension rods may be designed for 
the force 15,800 lbs. 

The forces sustained by the pillar are as follows (see Fig. 
276): FE, the horizontal component of ^£, = 15,000 lbs., acts 



MACHINE FRAMES. 



473 



horizontally toward the right at K, and EF, the horizontal 
component of EB, = 15,000 lbs. acts toward the left at H. BC = 
30,000 lbs. acts toward the left at H, and DA =30,000 lbs. acts 
toward the right at L. CD acts upward at L, producing a total 
compression in the portion LH of 10,000 lbs. The force AF = 
5000 lbs. acts to produce tension between H and K. From 
these data the pillar may be designed by methods already given. 




Fig. 276. 



261. Frame of a Steam Riveter. — Let Fig. 277 represent a 
steam riveter. Both the frame and the stake are acted upon by 
three parallel forces when a rivet is being driven. The lines of 
action of these forces AB, BC, and CA, are ab, be, ca. The 
force AB required to drive the rivet = 35,000 lbs. BC and CA 
may be found, the distances EH and HG being known. The 
moment of flexure on the line ca = 35,000 lbs. X 74" = 2,590,000 
inch-pounds. Let the line HF represent this moment. The moment 



474 



MACHINE DESIGN. 



in any horizontal cross-section may be found from the diagram 
EFG. Any section of the frame or stake may therefore be checked. 
The stake needs to be small as possible in order that small boiler 
shells and large flues may be riveted. In order that it may be 
of equal strength with the cast iron frame, it is made of material 
of greater unit strength, as cast steel. 




72000^ 



Fig. 277. 



The two bolts which hold the frame and stake together sus- 
tain a force of 107,000 lbs. The force upon each therefore is 
53,500 lbs. If the unit strength of the material is 50,000 lbs., 
and the factor of safety is 6, the area of cross -section of each bolt 

6x 53S°° 

would be= =6.42 square inches. The diameter corre- 

50000 

sponding = 2.86 inches. A 3^-inch bolt has a diameter at the 



MACHINE FRAMES. 475 

bottom of the thread = 2.88 inches, and and hence 3|-inch bolts 
will serve as far as strength is concerned. But the body of the 
bolt is 60 inches long, and each inch of this length will yield 
a certain amount, and the total yielding might exceed an 
allowable value, even if a safe stress were not exceeded. The 
yielding per inch of length, or the unit strain = unit stress -r- 
coefficient of elasticity, or 

1=^, but / = ^ =^ = 6440 lbs. 

and £ = 28,000,000 

6440 . 

. . a = —z = .00002^ men. 

28000000 ° 

Total yielding = X X 60 = .00138 inch. 

This amount of yielding is allowable and therefore two 3j-inch 

bolts will serve. 



APPENDIX. 



The following method of determining the position of the 
slider-crank chain corresponding to the maximum velocity of the 
slider is largely due to Professor L. M. Hoskins. 

Refer to Fig. 278. 




Fig. 278. 

BM = b = connecting-rod length, to scale. 

OM = a = crank length, to scale. 

OA=y. 

AM = x. 

OB = d. 

Angle AOB =90°. 

" MAO = a. 

" AMO = r . 

11 OMB = p. 
L = 2a = length of stroke of slider. 

From our study of the velocity diagram of the slider-crank 
chain (see § 22) we know that the length y will represent the 

477 



478 APPENDIX. 

velocity of the slider on the same scale as the length a repre- 
sents the velocity of the center of the crank-pin. The length 
y is determined by erecting at O a perpendicular to the line of 
action of the slider and cutting this perpendicular by the con- 
necting-rod b, extended if necessary. 

Our problem, then, is to find the position of the mechanism 
corresponding to the maximum value of y. 

Consider the triangle whose sides are y, x, and a. Calling 
the angle included between x and y, a, 

a 2 =y 2 +x 2 — 2xy cos a; (i) 



y 

but cosa' = — -t (2) 

x + b v ' 



/. a 2 =y 2 + x 2 -2xy—^ (3) 



Clearing (3) of fractions and transposing, 

2xy 2 = (x 2 + y 2 — a 2 ) (x + b), 
2xy 2 =x 3 +x 2 b+y 2 x+y 2 b—a 2 x—a 2 b. ... (4) 

Differentiating, 



dy dy dy 

2y 2 +4xy-~ =3X 2 + 2bx+y 2 + 2xy-r- +2by-j a 2 . 



Transposing, 



.dy 
(4xy — 2ccy — 2by)-r- =3X 2j r2bx+y 2 —a 2 — 2y 2 . . . (5) 



APPENDIX. 



479 



dy 
For maximum value of y, -r- =o; hence we may write o for 

the left-hand term of (5). 

o = 3# 2 + 2 bx + y 2 — a 2 — 2y 2 ; 
•"• y 2= 3x 2 + 2bx — a 2 (6) 

Adding x 2 and subtracting a 2 from both sides of (6), 

x 2 +y 2 — a 2 =4X 2 + 2bx — 2a 2 (7) 



From (3), 

x + b' 



2,2 2 2x y 

x 2 +y 2 — a 2 



Substituting this value in (7), 

2XV 

— ~r=AX 2 + 2bx — 2a 2 : 
x + b ' 

.". 2^ 2 = (4X 2 + 2&x — 2a 2 ) (x + b) (8) 

Substituting in (8) the value of y 2 given in (6), 

2#(3# 2 -f 2 bx — a 2 ) = (4JC 2 + 2 foe — 2 a 2 ) (# + &) ; 
.*. 6X 3 + ^bx 2 — 2a 2 x = 4.x 3 + /[bx 2 + 2bx 2 + 2b 2 x — 2a 2 x — 2ba 2 ; 
2X 3 — 2bx 2 — 2b 2 x + 2ba 2 =0; 
x 3 — bx 2 — b 2 x + ba 2 = o (9) 

Dividing (9) by a 3 and transposing, 

a 2 x x 2 x 3 

b 2= b + b 2 ~b 3 ( Io) 

Equation (10) gives us the relation existing between a, b, 
and x for the maximum velocity of the slider. 



480 APPENDIX. 

x 
By taking a series of values of 7- and solving (10) for the 

a 
corresponding values of j- } Curve A has been constructed. Ordi- 

oc . . a 

nates are 7/, abscissae are ■=-. 
b b 

For any given problem the values of a and b are known. 
Solve for 7-. 

From Curve 4 find the value of 7- corresponding to this value 

AT 

From the determined value of 7- and the known value of b 



the numerical value of x is found. 

But equation (6) gives for the maximum value of y the rela- 
tion 

y 2 =^x 2 + 2bx — a 2 t 

which we can readily solve for y since all of the right-hand terms 
are now known. 

AOB being a right-angled triangle, 

d 2 = (x + b) 2 -y 2 . 

The values of the right-hand member being known we can 
readily solve this for d % 

Let m represent the distance moved through by the slider 
from the beginning of the stroke, then 

m = b + a—d. 



APPENDIX. 4 8 x 

The portion of the stroke accomplished by the slider at the 
time of its maximum velocity expressed as a fraction of the 
whole stroke, 2 a, 

_m 
2a 



Curve B shows the relation between =- and — . From this 

2a 

curve we can see at a glance for any given value of r what per 

cent of the slider's stroke is accomplished when its position of 

a 
maximum velocitv is reached. Abscissae are values of 5-; ordi- 



m 
nates — . 
2a 

y 

— = ratio of the maximum velocity of the slider to the velocity 

of the center of the crank-pin. Curve C shows the relation 

ay a 

between the values of 7- and — . Abscissae are values of t ; ordi- 
b a b ' 

y y 

nates - — 1. Add unitv to the ordinates for actual values of -. 
a J a 

To find the values of /? corresponding to the maximum velocity 

of the slider we have the three sides of the triangle OMB, namely> 

b, a, and d. Let 

b+a+d 
s= . 



Then cos J/9 = — 7 — - , from which we can readily get the 

value of t 'i. 

Curve D is plotted with values of /?, in degrees, as ordinates 

a 
and values of r- as abscissae. 




482 



APPENDIX. 



Table XXXI. — Assumed Values or — , and Corresponding Computed Values 

b 

a y m y 

OF T. T", , — , AND B, TO PLOT CURVES A, B, C, AND D. 

o b 2a' a r 



X 

b~ 


O.OIO 


0.015 


0.025 


0-035 


0.050 


0.075 


O.IOO 


0.200 


a 
~b 


0.1005 


0.1234 


0. 1600 


0. 1902 


0.2289 


0.2832 


0.3302 


0.4817 


y 

b 


O. IOI 


0.1243 


0. 162 1 


0.1936 


0.2348 


. 2944 


0.3479 


0-5367 


m 

2a 


Q-475 1 


0.4704 


0.4622 


0.4561 


. 4484 


. 4402 


0.4320 


0.4239 


a 


1.005 


1.0073 


1.0131 


1 .0179 


1.0258 


i-o395 


1-0533 


1.1139 


fi 


89° 55' 


8o° 49' 


8o° 44' 


8 9 ° 35' 


89 26' 


88° 50' 


88° 16' 


85° 1/ 



0.300 

0.6025 

0.7121 

0.4273 

I. 1819 
8i° 23' 



0.400 
o - 7043 
0.8854 
o . 4409 

I. 2571 

76° 41' 



0.500 

o . 7906 
1.0607 

o-4597 
1. 3416 



0.600 
0.8626 

i- 2 393 
0.4901 

1-4367 
65 19' 



o. 700 

0.9203 

1.4223 

0-5374 

1-5455 



0.800 
0.9633 
1 . 6100 
0.6044 

1-6713 
4 8° 21' 



0.900 

0.9905 
1.8023 
0.7221 

1. 8198 
35° 8' 



APPENDIX. 



483 




INDEX. 



PAGE 

Accelerated and retarded motion, cams for 54 

Acceleration, diagrams of 68, 75, 79 

General method diagram 79 

Adaptation in design ix 

Addendum 349, 359 

Ahara, E. H 316 

Alford's Bearings 244, 247 

American Journal of Science 213 

American Machinist 122, 125, 173, 179, 217, 229, 266, 269, 280, 321, 

344, 372, 412,432, 442 
American Society of Mechanical Engineers. 

See Journal; Transactions; also Boiler Code. 

Andrews, E. S 442 

Angle of action 356 

Annular gears 362 

Appearance in design xii 

Application, machinery of 4 

Arc of action 356, 374, 375 

Archbutt and Deeley's Lubrication and Lubricants 252 

Axle design 194 

Axles, shafts and spindles 194-209 

Bach, Prof. C 104, 124, 198, 284, 359, 378, 380 

Bach, Prof. C. and E. Roser 416 

Ball Bearings. See Roller- and Ball-bearings. 

Barlow's cylinder formula , 103 

Barnard, \V. N 1 20 

Barr, John H 120, 191 

B arth, Carl G 301 , 344 

B IUSCHINGER, J 83 

Beams, Tables of 102 

Bearing Pressure, allowable for journals 230, 231, 242, 244, 247 

Allowable for roller- and ball-bearings 269 

Allowable for sliding surfaces 19 1 

Allowable for thrust-bearings 242, 244, 247 

48S 



486 INDEX. 

PAGE 

Bearings 247-260 

See also Journals; Roller- and Ball-bearings. 

Belts 286-310, 321 

Coefficient of friction of 301 

Cone or stepped pulleys for 292 

Creep and slip of 301 

Crowning pulleys for 292 

Design, theory of 295 

Distances between shafts for 310 

Driving capacity, variation of 307 

Dynamo-belt design 305 

Efficiency of 302 

Intersecting axes 290 

Pump-belt design 304 

Shifting, principle of 289 

Size of pulleys for 310 

Steel 321 

Transmission of motion by 286 

Twist . 289 

Weight of leather 300 

See also Chains; Rope transmission. 

Bending moments of beams 102, 103 

Benjamin, Prof. C. H 343 

Bevel-gears 382 

Skew 389 

Bird, Prof. W. W 301 

Birnie's cylinder formula 103 

Boiler code 112. 129, 138 

Shell, design of 131 

Bolts and screws 139-167 

Analysis of screw action 142-145 

Calculation of bolts subject to elongation 152 

Calculation of screws for transmission of power 160 

Calculation of screws not stressed in screwing up 145 

Calculation of screws stressed in screwing up 146 

Cap screws 140 

Classifications and definitions 139-150 

Coefficient of friction of 162 

Design of bolts for shock 156 

Efficiency of 162 

Lock-nuts 159 

Set-screws 140, 177 

U. S. standard threads 140 

Wrench pull 151 

Box pillar. See Supports. 



INDEX. 487 

PAGE 

Boxes 247-260 

See also Journals. 

Brackets 95 

See also Supports. 

Brakes, band and block 323 

Brass 88, 99 

See also Boxes; Bronze. 

Bronze, manganese, naval phosphor, and Tobin 88, 99 

See also Gun-metal. 

Brown & Sharpe Manufacturing Co 364, 366, 369, 374, 398, 413 

Brown Hoisting and Conveying Machine Co 322 

Browne, D. H 200 

Bruce, R. A 417 

BlTHNER, K 378 

Cams 51-61 

Carpenter, R. C 158 

Cassier's Magazine 342 

Cast-iron 88, 91, 98, 99 

Parts 96 

Centro n-13 

Location of 16-17, 18, 22-24 

Centrode 13-14 

Centros of three links 1 7-18 

Chains, Block 326 

Efficiency of 326 

Flat link 326 

Morse 327 

Motion 15 

Open link 325 

Renold 327 

Roller 326 

Stud link 325 

Church, Prof. I. P 239 

Circular plates, stresses in 104 

Clauzel, M. Le Baron 121 

Claverino's cylinder formula 103 

Clearance, Ball-bearing 272 

Gear tooth 349 

Clutches 279-285 

See also Couplings. 

Columns, long, formulse 103, 200 

Cos ant, D.J ill 

Connecting-rod, angularity of 30-32 

Constrained motion 5-8 



488 INDEX. 

PAGE 

Copper 88, 98, 99 

Cotters 176 

Couplings and clutches 274-285 

Band 284 

Claw, jaw, or toothed 279 

Combination, friction and claw 284 

Compression 275 

Defined 274 

Disengaging 279-285 

Flange 275 

Flexible 278 

Friction 280 

Hall's 278 

Hooke's 277 

Hydraulic 285 

Magnetic 285 

Oldham's 277 

Permanent 274-279 

Pneumatic 285 

Self-sustaining 282 

Seller's _ . 276 

Sleeve, muff, or quill 274 

Weston friction • 283 

Cranes, problems in design of 461-473 

Crank-pin, design of 232, 236 

Cresson, The Geo. V. Co 313 

Critical speed of shafts 204 

Cross-head pin, design of . 237 

Cross-head, slotted 20-21 

Curtis turbine step-bearing 244 

Cutter, L. E hi, 374 

Cutting Speeds 32 

Cycloidal gears 353~358, 3 6 °> 373, 3%3 

Cylinder formulae 103 

Heads, stresses in 104 

Deflections, beams 102 

Line-shafts 203 

Dewrance, J 224, 254 

Dynamo, belt-design for 288, 305 

Economy in design x-xii 

Efficiency of machines 2 

Elastic limit, table of values 99 

Elasticity, moduli of 99 



INDEX. 489 

PACE 

Electric conductivity 98 

Elements, pairs of motion , 14 

Size oT 19 

Elliptical plates, stresses in 104 

Energy, definition of 1 

In machines 2, 62-80 

Law of conservation of 1 

Sources of 3 

Engineering (London) 180 

Engineering News 285 

Engineering Record 247 

Ewixg, Prof. J. A 338 

Factor of safety, discussion of 82 

For gear teeth 374 

Fastenings. See Riveted Joints; Bolts and Screws; Keys; Cotters; Fits 

and Fitting. 
Feathers. See Keys. 

Fellow's stub-tooth gears 375 

Fits and Fitting 169, 178-183 

Flanged plates, stresses in 104 

Flat plates, stresses in 104 

Flather, J.J 3n,3i5, 316 

Fly-wheels 328-345 

Construction of 341 

Design, general method 329 

Haight's joint for 345 

Pump 333 

Punching machine 329 

Steam-engine 336 

Stresses in arms 339 

Stresses in rims 336 

Theory of 328 

Followers 51-61 

Force, definition of 1 

Force-fits 178 

Form, dictated by stress 81, 89 

Frames 92-95, 44i"475 

Closed 452 

Cranes 461 

Open-side 44 r 

Punching-machine 442 

Riveting-machine 473 

Slotting-machine 447 

Steam-engine, center-crank 452 



49° INDEX. 

PAGE 

Frames {continued). 

Steam-engine, side-crank 449 

Steam hammer 455 

See also Supports; Machine parts. 

Friction brakes 323 

Clutches 280 

Coefficient for belts 302 

brakes and clutches 282 

cotters 177 

dry surfaces 212 

force fits 183 

lubricated surfaces 213, 216 

roller- and ball-bearings 273 

ropes 311 

screws 162 

wire ropes 320 

Energy loss by 2 

Gear tooth 378 

Heat generated by, in journals 220 

Laws of, dry surfaces 212 

, imperfectly lubricated surfaces 213 

, perfectly lubricated surfaces 214 

See also Journals; Lubrication; Sliding Surfaces. 

Fritz, John 343 

Gearing. See Toothed Wheels. 

General Electric Co 230 

Gleason, Andrew 376 

Works 386 

Grant's Teeth of Gears 369 

Graphite, as lubricant 259 

Grashof's cylinder formula 103 

Grasshopper motion 4? 

Green, B. M iii 

Greenhill, A. G 201 

Guest's maximum shear theory 198 

Guides, see sliding surfaces 185-193 

Gun-metal 88, 99 

Halsey, F. W 179, 390, 398, 432, 435 

Harmonic motion, cams for 54 

Hess, Henry 270 

Higher pairs 14 

Hill, T 266 

Hirn, G 219 



INDEX. 491 

PAGE 

Hoskins, Prof. L. M 445, 452, 477 

Hubs, size of 344, 370 

Stresses in 179, 183 

Hunt, The C. \V. Co 315 



Illinois, University of, Bulletin 127, 173, 325 

Indicator pencil njechanisms 42-46 

Crosby 46 

Tabor 45 

Thompson 44 

Instantaneous center 10-1 2 

Instantaneous motion 10 

General solution for 30 

Involute gears 358, 361, 373, 383 

Iron. See Cast-iron; Wrought-iron. 



Janney, R 269 

Jib-crane 461 

Johnson, Bryan, and Turneaure's Theory and Practice of Modern Framed 

Structures 467 

Jones' Machine Design 344 

Journal, Amer. Soc. of Mech. Engs 205 

Amer. Soc. of Naval Engs 126 

Franklin Institute 81 

Journals 210-260 

Allowable bearing pressure 230, 231, 242, 244, 247 

Calculation of, for strength 232 

Crank-pin of engine 236 

Cross-head pin of engine 237 

Design of, by heat balance 224 

Friction of 211 

General discussion of 210 

Heating of 2 20 

Lasche's experiments on 219 

Lubrication of 211, 252 

Main, of engine 233 

Materials for, and bearings 231 

Moore's experiments on 217 

Proportions of 231 

Stribeck's experiments on 219 

Thrust 239 

Tower's experiments on 217 

See also Roller- and Ball-bearings. 



492 INDEX, 

PAGE 

Kenerson, W. H 417 

Kennedy, Sir A. B. W 13, 19, 50, 107, 427 

Keys 168-177 

Classification of 168 

Cotters 176 

Feathers 174 

Kernoul and Barbour 172 

Parallel 168 

Roller ratchet 1 73 

Round taper 175 

Saddle, flat and angle 172 

Splines 1 74 

Strength of 1 73 

Taper 169 

Woodruff • 170 

Keyways, effect of, on strength 1 73 

Kimball, A. S 213 

Kingsbury, Albert 161, 193, 246 

Lanza, Prof. G 177, 213, 301 

Lasche, W 219, 220, 223, 378, 379 

Lathe, bed 95, 188 

Supports 437 

Le Conte, J. N 72, 398 

Legs. See Supports. 

Lever-crank chain, location of centros 19 

Velocity diagram of 28 

Lewis, Wilfred '. 213, 296, 301, 372, 373 

Linear velocity 24 

Points in different links 28 

Line-shafts v 202 

Linkage, definition of 15 

Compound 16 

Simple 16 

Lock-nuts 159 

Lof, E. A 209 

Long columns 90, 103, 200 

Lower pairs 14 

Lubrication of helical surfaces 162 

of roller- and ball-bearings 272 

of rotating surfaces 211, 252 

of sliding surfaces 192 

McCord's Kinematics 369 

Machine, cycle 2 



INDEX. 493 

PAGE 

Machine, definition of i 

Efficiency of 2 

Frames, dictated by stress 9 2_ 95 

See also Frames. 

Function of 3 

Parts, forms of cast members 96-97 

Parts, proportions dictated by stress 81-104 

Machinery 209, 376 

Main journal, design of 233 

Master Steam Boiler-makers Assn in 

Materials, Tables of properties of 88, 98-104 

Maw's Modern Practice in Marine Engineering 252 

Mechanism, definition of 16 

Location of centros in compound 22 

Mechanisms, quick-return 32-40 

Melting points of metals 98 

Merrimax, Prof. M 199, 200 

Metals, tables of physical properties of 98 

Miller, Spexcer 313 

Mitchel, A. G. M 246 

Modulus of elasticity 99 

of rigidity 99 

of rupture 99 

of section, plane and polar 100-101 

Moments of inertia ioc—ioi 

Moore, Prof. H. F 127, 173, 217 

Morley, Prof. A 180 

Morse Chaix Co 327 

Moss, Saxford A 103, 183, 338 

Motion, chains 15 

Constrained 5-8 

Definition of 1 

Elements 14, 19-20 

Free 4 

Helical 9 

Instantaneous 11, 30 

Kinds of, in machines 8-9 

Plane 8 

Relative 9 

Spheric 9 

Pairs of elements 14 

Parallel motions. See Straight-line motions 41-50 

Parallelogram 41 

Passive resistance 5-8 



494 INDEX. 

PAGE 

Peaucellier link 48 

Peddle, Prof. J. B 432 

Pederson, Axel 229 

Petroff's Neue Theorie der Reibung 216 

Philosophical Transactions 216, 252 

Pillar Crane 472 

P| tch arc ; 356, 374, 375 

Pitch, axial 417 

Circular or circumferential 349 

Diametral 349 

Normal 391 

Pitch circle 348 

Planing-machine, bed 94 

Lubrication of 193 

Table 93 

Plates, stresses in flat, flanged and stayed 104 

Plymouth Cordage Co 315 

Poifson's ratio, table of values of 99 

Porter, C. T 232 

Power 278, 344 

Pratt, C. R 266, 271 

Pressure-vessel walls, formulae for 103-104 

Prime mover 3 

Proceedings Amer. Soc. Civil Engs 313 

Inst. Civil Engs 224, 254 

Inst. Mech. Engs 107, in, 117, 121, 217, 240, 242, 252, 255, 417 

Master Car Builder's Assn 196 

Phila. Engs. Club 372 

Pulleys, cone 292 

Crowning 292 

Idler or guide 291 

Proper size of, for belts 310 

Proportions of 344 

Stresses in arms 339 

Stresses in rims 336 

See also Fly-wheels. 

Pump, belt design for 304 

Fly-wheel design for 333 

Punching machine, fly-wheel design for 329 

Frame design for 442 

Quick-return mechanisms 32-40 

Lever-crank quick -return 34~37 

Slider-crank quick-return 32-34 

Whitworth quick-return 37-40 



INDEX. 495 

PAGE 

Radii of gyration, table of ioo 

Railway Machinery 196 

Rankine's Machinery and Millwork 50 

Rectangular plates, stresses in. . . 104 

Repetitive stresses. See Stresses, variable. 

Reuleaux's Constructor 196, 241, 315, 318, 390, 408, 435 

Reynolds, Prof. Osborne 216, 252 

Richards, John 81, 92, 168, 175 

Rigid body 10 

Rigidity, modulus of 99 

Risdon Iron Works 172 

Riveted joints 105-138 

Boiler-shell problem 131-138 

Countersunk rivets 126 

Dimensions of rivet heads 125 

Efficiencies of 113—121, 123 

Failure of no 

General formulae for 1 20 

Kinds of 108 

Length of rivet 125 

Margin 122 

Materials for 1 29 

Methods of riveting 105 

More than two plates ' 130 

Xickel-steel rivets 126 

Perforation of plate 106 

Plates not in same plane 131 

Plates with upset edges 129 

Slippage of 1 24 

Strength of materials used in 111-113, 126, 129 

Strength, proportions and efficiency of 1 13-123 

Tightness of 122, 124, 127 

Riveting-machine, action of 105 

Frame design for 473 

Robinson, Prof. S. W 369 

Roller- and ball-bearings 261-273 

Allowable loading for 269 

Binding, prevention of 273 

Efficiency of 273 

General considerations 261 

Lubrication and sealing of 272 

Races for 261, 268 

Rolling, sliding, and spinning 261 

Size of 271 

Rope-transmission, fibrous 310 



496 INDEX. 

PAGE 

Rope-transmission, wire 318 

Rupture, modulus of 99 

Safety, factor of 82 

of machine operators xii 

Sames, CM... 103 

Screws and screw threads. See Bolts and screws. 

Section modulus, tabe of plane and polar 100- 101 

Set-screws 140, 177 

Shafting, angular distortion of 200, 202 

Combined thrust and torsion 200 

Combined torsion and bending 198 

Critical speed of 204 

Hollow vs. solid 199 

Line-shafts 201 

Simple torsion 197 

Shaping-machine, force problem 64 

Quick-return mechanism for ^7 

Sheave-wheels, for fibrous ropes 312 

for wire ropes 320, 322 

See also Fly-wheels. 

Shibata, M 125 

Shrink-fits. • 178-183 

Sibley College 'Laboratories 158 

Sibley Journal of Engineering 1 20 

Skew bevel-gears 389 

Slider-crank chain, acceleration diagrams 68-75, 79 

Description of 16-17 

Force problem, shaping-machine 64-66, 75 

Force problem, steam engine 66-74, 76-79 

Location of centros 16-1 7 

Maximum velocity of slider 69, 477-483 

Tangential effort diagrams 76-78 

Velocity diagram 26, 28, 79 

Sliding pair 14 

Sliding surfaces 185-193 

Allowable bearing pressure for 191 

Form of guides of 187 

General discussion of 185 

Lubrication of 190 

Proportions dictated by wear 185, 193 

Slotted cross-head 20-21 

Slotting-machine, frame design for 447. 

Gearing for 380 

Smith, C. A 293 



INDEX. 497 

PAGE 

Smith, Oberlin 230 

Smith's Materials of Machines 81 

Souther, H 280 

Specific gravity of metals 98 

Spheres, stresses in 103 

Spindles. See Axles, shafts, and spindles. 

Spiral gears 390-41 2 

Axes at 90 390 

Axes at any angle, 405 

Splines. See Keys. 

Springs 429-435 

Axial or helical 433 

Cantilever 430 

Coil or spiral 434 

Elliptic and semi-elliptic 43 2 

Flat 430 

Leaf 431 

Materials and stresses 434 

Torsional 434 

Spur-gears 347-380, 421 

Square plates, stresses in 104 

Stayed surfaces, stresses in 104 

Steam-engine, boxes ". 249 

Crank-pin 232, 236 

Cross-head pin 237 

Fly-wheel design 336 

Force problem 66-74, 76-79 

Frame, center crank 452 

, " girder bed " 449 

, " heavy duty " 451 

Main journal 233 

Steam-hammer, double-acting, frame design 455 

Single-acting, frame design 455 

Steels, properties of various 88, 98, 99, 112, 434. 

Stodola's Sleam Turbines 22>\ 

Stoney's Strength and Proportions of Riveted Joints 107, in, 118, 123 

Straight-line motions 41-50 

General methods of design 44-48 

Grasshopper 43 

Parallelogram 41 

Peaucellier 48 

Watt 41 

Strength and stiffness in design ix 

See also Stresses. 

Stress and strain formulae 102-104 



498 INDEX. 

PAGE 

Stresses in machine parts, combined 198, 200 

Compression 89 

Constant 82 

Flexure 90 

Shock or suddenly applied 87 

Tables of 102-104 

Tension of 89 

Torsion 91 

Variable 82-87 

Stribeck, Prof. C 217, 218, 219, 220, 223, 224, 269, 273, 416. 417 

Stud, definition of 139 

Supports 436-446 

Divided 437 

General laws for design of 436 

Reduced number of 439 

Three-point 438 

See also Brackets; and Frames. 
Sweet, Prof. John E v, 81, 94, 158, 187, 439, 440 

Talbot, Prof. A. N 127 

Taylor, Frederick W 301 

Thomas, Prof. Carl 273 

Thomas's Worm Gearing 390 

Thrust-journals 238-247 

Thurston, Prof. R. H 219 

Toothed wheels or gears 347-428 

Addendum 349, 359 

Angle of action 356 

Annular 362 

Arc of action 356, 374, 375 

Backlash 349 

Bevel-gears 382 

Bronze 379 

Cast-iron 374, 375, 379 

Circular-pitch 349 

Clearance 349 

Cycloidal teeth 353~358, 360, 373, 383 

Depth, total 349 

Depth, working 349 

Diametral pitch 349 

Efficiency 379 

Elliptic 380 

Epicyclic trains 424 

Forms of teeth 350, 366 

Friction, pressure and abrasion 378 



INDEX. 499 

PAGE 

Toothed wheels or gears {continued). 

Hard fiber 376 

Helix angle 390 

Interchangeable sets . . . .' 354, 363, 365 

Interference 359 

Involute teeth 358, 361, 373, 383 

Line of pressure 356 

Non-circular wheels 380 

Pinion 360 

Pitch arc 356, 374, 375 

Pitch circle 348 

Proportions of 369 

Racks 360 

Rawhide 376, 379 

Reverted trains 423 

Skew-bevel 389 

Spiral 390 

Spur gear-chains 421 

Spur wheels 347-380 

Steel 376, 379 

Step, twisted or herring-bone 381 

Strength of teeth 371 

Stub-tooth 375 

Theory of 347 

Twisted bevel 388 

Velocity coefficients 374, 375 

Worms and wheels 412 

Torque diagrams 76-78 

Tower, Beauchamp 217, 219, 240, 241, 242, 243, 252, 253, 254, 255 

Traction and Transmission 219 

Transactions, Amer. Inst. Min. Engs 200 

Transactions, Amer. Soc. Mech. Engs 142, 161, 177, 179, 183, 191, 

210, 213, 230, 232, 242, 254, 271, 273, 280, 293, 301, 316, 336, 338, 

343, 344, 374, 393, W, 435 

Transmission, machinery of 4 

See also Belts; Ropes; Shafting; and Toothed Wheels. 

Turning pair 14 

Twisting pair 14 

Unwtn's Machine Design 84, 87, 130, 191, 219, 413 

Value of metals, approximate 98 

Vector quantity 25 

Velocity, angular 24 

Definition of 24 



5 oo INDEX. 

PAGE 

Velocity, angular {continued). 

Diagrams 26, 28, 79 

Linear 24 

Relative ' 24, 26, 28 

Watertown Arsenal, Tests of Metals 107, in, 123 

Watt parallel motion 41 

Ways. See Sliding Surfaces. 

Weaver, S. H 205 

Weight of Metals 98 

Weisbach's Mechanics of Materials 50 

Weyrauch, J 83 

White, Maunsel 107 

Whitworth quick-return mechanism 37 

Willis's Elements of Mechanism 369 

WOHLER, A 8^ 

Work, definition of 1 

Worm-gearing 41 2 

Wrought-iron 88, 98, 99 

Zeitschrift des Vereins deutscher Ingenieure. . 217, 219, 269, 273, 378, 379, 416, 417 
Zeitschrift fur Math, und Physik 246, 247 






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